Calculus @ MindSay

This is just a good ole summary section.  Nothing new.  Now you can just get all your main equations from the chapter in one nice easy section!

Fundamental Theorem of Calculus:

∫[a,b] (F'(x)dx) = F(b) - F(a)

Produces a distance between two values, aka a straight line between values at a and b.  Ultimate relates an integral to function values.

Fundamental Theorem for Line Integrals

∫[C] (Ñf ∙ dr) = f(r(b)) - f(r(a))

Produces an arc length between two values, aka a curve C between values at r(a) and r(b).  Ultimately relates a line integral to values of vector-valued functions.

Green's Theorem

∫∫[D] ((∂Q/∂x - ∂P/∂y)dA) = ∫[C] (Pdx + Qdy)

Produces the area of a region D bounded by a curve C, taking counterclockwise as positive orientation.  Ultimately relates a double integral to a line integral.

Stokes's Theorem

∫∫[S] (curl F ∙ dS) = ∫[C] (F ∙ dr)

Considers only a surface (positive oriented with n outward) and a boundary curve on some area of this surface and produces the area of this section.  Ultimately relates a surface integral to a line integral.

Divergence Theorem

∫∫∫[E] (div F dV) = ∫∫[S] (F ∙ dS)

Produces a volume of a surface given by an enclosed region E, positive oriented with n outward.  Ultimately relates a triple integral to a surface integral.

Earlier, we said Green's Theorem could be written in a vector form as

∫[C] (F ∙ n ds) = ∫∫[D] (div F(x,y) dA)

For giggles, is it possible that we could create a similar set up for a surface integral?  Why yes!  Yes we can!  It works for a special type of region, called a simple solid region.  In Chapter 16, in triple integration, we had three types of regions.  Well, if you take something like, say, a rectangular box, that can be ANY of the three types of regions, all of your choosing, and all at once.  Such a region is a simple solid region if it can be all three types of regions at once.  Also, we chose positive orientation for surfaces to be outward, that is, n, the unit normal vector, points outward, not inward.  With those two in mind, we can now create the Divergence Theorem:

Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation.  Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E.  Then

∫∫[S] (F ∙ dS) = ∫∫∫[E] (div F dV)

And we'll prove it at the end.

And that is IT.  We'll do an example.

We wanna evaluate ∫∫[S] (F ∙ dS), where F(x,y,z) = xyi + (y² + e^{xz^2})j + sin(xy)k, and S is the surface of the region E bounded by the parabolic cylinder z = 1 - x² and the planes z = 0, y = 0, and y + z = 2.

Normal methods of surface integrals would be horrendous.  Not only would we have to evaluate FOUR surface integrals (because this solid region has four surface boundaries), but look at F, and you see it's sort of a nightmare.

However, we'll take div F:  ∂[xy]/∂x + ∂[y² + e^{xz^2}]/∂y + ∂[sin(xy)]/∂z
= y + 2y = 3y

So, instead of some "mess" for F ∙ dS, we have div F dV, which would be 3ydxdydz, though not necessarily the "d_" in that order.  Let's look at our surface.

If we say E is a type 3 region, then we have E = {(x,y,z) | -1 < x < 1, 0 < z < 1 - x², 0 < y < 2 - z}, and so we finally wind up with

∫∫[S] (F ∙ dS) = ∫∫∫[E] (div F dV) = ∫∫∫[E] (3ydV)
= 3*∫[-1,1]∫[0,1-x²]∫[0,2-z] (ydydzdx)

Integrating to y, we get y²/2 from 0 (trivial) to 2 - z, which is just (2 - z)²/2.  To integrate that, we let u = 2 - z, du = -dz, so now we have (-3/2)*stuff (we pull out the 1/2), so we have u² in there and that integrates to u³/3, or (2 - z)³/3 from 0 to 1 - x², which is just (1 + x²)³/3 - 2³/3 = (1 + x²)³/3 - 8/3, or (1/3)[(1 + x²)³ - 8], which simplifies to x⁶ + 3x⁴ + 3x² + 1 - 8, or

(-1/2)*∫[-1,1] ((x⁶ + 3x⁴ + 3x² - 7)dx)

Integrating to x gives x⁷/7 + 3x⁵/5 + x³ - 7x from -1 to 1.  That's (1/7 + 3/5 + 1 - 7) - (-1/7 - 3/5 - 1 + 7) = (2/7 + 6/5 + 2 - 14) = -368/35.  Multiply that by -1/2 from earlier, and you get 184/35.  The end.

At the end, this theorem is proved for simple solid regions, but it can also be proved for finite unions of simple solid regions (like, say, an egg within a ball, for example).

Suppose E lies between the closed surfaces S₁ and S₂, where S₁ lies inside of S₂.  Let n₁ and n₂ be outward normals of S₁ and S₂.  Then the boundary surface of E is S = S₁ unioned with S₂, and its normal n is given by n = -n₁ on S₁ and n = n₂ on S₂.  Applying the Divergence Theorem,

∫∫∫[E] (div F dV) = ∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS)
= ∫∫[S₁] (F ∙ (-n₁) dS) + ∫∫[S₂] (F ∙ n₂ dS)
= -∫∫[S₁] (F ∙ dS) + ∫∫[S₂] (F ∙ dS)

To continue, let's pull an application.  Suppose we have an electric field E(x) = εQx/|x|³, where S₁ is a small sphere with radius a and center at the origin.  div E = 0, and so we have

∫∫[S₂] (E ∙ dS) = ∫∫[S₁] (E ∙ dS) + ∫∫∫[E] (E ∙ n dS) from a rearrangement of the above mess.  The normal vector at x is x/|x|, so

E ∙ n = εQx/|x|³ . (x/|x|) = εQx/|x|⁴  ∙ x
= εQ/|x|² = εQ/a² because the equation of S₁ is simply |x| = a.  And so, we have

∫∫[S₂] (E ∙ dS) = ∫∫[S₁] (E ∙ n dS)
= (εQ/a²)*∫∫[S₁] (dS) = (εQ/a²)*A(S₁)
= (εQ/a²)4πa² = 4πεQ

So, the flux of E is 4πεQ through ANY closed surface S₂ that contains the origin.

Another application comes from fluid flow.  Let v(x,y,z) be the velocity field of a fluid with constant density ρ.  Then F = ρv is the rate of flow per unit area.  If P₀(x₀,y₀,z₀) is a point in the fluid and B_a is a ball with center P₀ and a VERY small radius a, then div F(P) ~ div F(P₀) for all points in B_a since div F is continuous.  We approximate the flux over the boundary sphere S_a as

∫∫[S_a] (F ∙ dS) = ∫∫∫[B_a] (div F dV)
~ ∫∫∫[B_a] (div F(P₀) dV)
= div F(P₀) *V(B_a), where V(B_a) is the volume of the ball.

This approximate becomes better as a -> 0, and so we can suggest that

div F(P₀) = lim (1/V(B_a))*∫∫[S_a] (F ∙ dS) as a -> 0

div F(P₀) is the net rate of outward flux per unit volume at P₀, and if div F(P) > 0, the net flow is outward near P and P is called a source.  If div F(P) < 0, the net flow is inward near P and P is called a sink.

Anyway, that's really it.  Let's do some problems!

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1)  Use the divergence theorem to prove that ∫∫[S] (D_nf dS) = ∫∫∫[E] (Ѳf dV), assuming that S and E satisfy the conditions of the divergence theorem and the scalar functions and components of the vector fields have continuous second order partial derivatives.

Not hard.  D_nf = Ñf ∙ n, so

∫∫[S] (D_nf dS) = ∫∫[S] (Ñf ∙ n dS) = ∫∫∫[E] (div (Ñf)dV) = ∫∫∫[E] (Ѳf dV)

2)  Use the divergence theorem to evaluate ∫∫[S] (F ∙ dS) where F(x,y,z) = x⁴i - x³z²j + 4xy²zk, where S is the surface of the solid bounded by the cylinder x² + y² = 1 and the planes z = x + 2 and z = 0.

Well, we need div F = 4x³ + 0 + 4xy², so we have div F = 4x(x² + y²).  If we look at the plane z = x + 2, we see an ellipse of x² + y² = 1, and we can convert this to cylindrical coordinates by saying that E = {(r,θ,z) | 0 < r < 1, 0 < θ < 2π, 0 < z < rcos(θ) + 2} by simply saying x = rcos(θ), y = rsin(θ), z = z, and so z = rcos(θ) + 2, x² + y² = 1 = r², r = 1.

That gives us ∫[0,2π]∫[0,1]∫[0,rcos(θ)] (4cos(θ)r²rdzdrdθ) when we convert 4x(x² + y²) into 4cos(θ)r².

So, we have to integrate 4cos(θ)r³dzdrdtheta with respect to z first.  That's just 4cos(θ)r³z from 0 (trivial) to rcos(θ), giving us 4cos²(θ)r⁴.  Integrating to r, we get (4/5)cos²(θ)r⁵ from 0 (trivial) to 1 gives us (4/5)cos²(θ).  cos²(θ) = 1/2 + cos(2θ)/2, so integrating THAT gives us θ/2 + sin(2θ)/4 from 0 (trivial) to 2π (trivial in the sine), giving us 2π/2 = π.  So, we have (4/5)*π = 4π/5.

3)  Use the divergence theorem to evaluate ∫∫[S] ((2x + 2y + z²)dS) where S is the sphere x² + y² + z² = 1.

We need to find some F ∙ n = 2x + 2y + z².  If we take the partial derivatives on S, we get 2x + 2y + 2z = 0, or x + y + z = 0, and we can turn that into our n via n = (the partial)/sqrt(sum of squares of partial), which ultimately amounts to (xi + yj + zk)/sqrt(x² + y² + z²), ahh, but we said the sphere has a radius of 1, so we can replace that bottom bit with just sqrt(1), giving us 1, so n = xi + yj + zk.

Now we have F ∙ (xi + yj + zk) = 2x + 2y + z².  Simple analysis will tell us that if we multiply the first two terms in xi + yj + zk by 2, we'd get our 2x + 2y.  Then if we multiply the last term by z, we'd get our z², so F = 2i + 2j + zk.

div F = 1, then, and if B = {(x,y,z) | x² + y² + z² < 1}, then we can easily say ∫∫[S] ((2x + 2y + z²)dS) = ∫∫∫[B] (1dV) = V(B) = (4/3)πr³ = 4π/3.

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Why do you care?  Well, the initial example gave us a MUCH easier way of evaluating surface integrals, for one.  The divergence theorem also allows us to analyze vector fields with more relative ease than just defining control surfaces/boundaries/volumes and integrating/evaluating from there.  You saw two applications in fluid mechanics and electricity right there.

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Proof of Divergence Theorem for simple solid regions.

Let F = Pi + Qj + Rk, then div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z

∫∫∫[E] (div F dV) = ∫∫∫[E] ((∂P/∂x)dV) + ∫∫∫[E] ((∂Q/∂y)dV) + ∫∫∫[E] ((∂R/∂z)dV)

If n is the unit outward normal of S, then the surface integral ∫∫[S] (F ∙ dS) is

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS) = ∫∫[S] ((Pi + Qj + Rk) ∙ n dS)
= ∫∫[S] (Pi ∙ n dS) + ∫∫[S] (Qj ∙ n dS) + ∫∫[S] (Rk ∙ n dS)

To prove the divergence theorem, we need to prove the following three equations:
∫∫[S] (Pi ∙ n dS) = ∫∫∫[E] ((∂P/∂x)dV)
∫∫[S] (Qj ∙ n dS) = ∫∫∫[E] ((∂Q/∂y)dV)
∫∫[S] (Rk ∙ n dS) = ∫∫∫[E] ((∂R/∂z)dV)

We'll do the last one and make a note that they can all be done similarly.

We'll let E be a type 1 region and so E = {(x,y,z) | (x,y) in D, u₁(x,y) < z < u₂(x,y)}, where D is the projection of E onto the XY plane.  So

∫∫∫[E] ((∂R/∂z)dV) = ∫∫[D] (Int[u₁,u₂] ((∂R/∂z)(x,y,z)dz)  dA), and by the Fundamental Theorem of Calculus:

∫∫∫[E] ((∂R/∂z)dV) = ∫∫[D] ((R(x,y,u₂) - R(x,y,u₁))dA)

Hold onto that.  Now let's look at the other part of the equation.  The boundary surface S consists of three pieces:  bottom S₁, the top S₂, and a possibly vertical S₃ that lies above the boundary curve of D.  If, say, S were a sphere, S₃ wouldn't even really exist.

On S₃, we have k ∙ n = 0, because k is vertical and n is horizontal, and so

∫∫[S₃] (Rk ∙ n dS) = ∫∫[S₃] (0dS) = 0

And so, regardless of what surface we have, we can say, then, that

∫∫[S] (Rk ∙ n dS) = ∫∫[S₁] (Rk ∙ n dS) + ∫∫[S₂] (Rk ∙ n dS)

S₂ is z = u₂(x,y), (x,y) in D, ad the outward normal n points upward, so

∫∫[S₂] (Rk ∙ n dS) = ∫∫[D] (R(x,y,u₂)dA).

On S₁, z = u₁(x,y), (x,y) in D, but here the outward normal n points downward, so we have

∫∫[S₁] (Rk ∙ n dS) = -∫∫[D] (R(x,y,u₁)dA)

Thus,

∫∫[S] (Rk ∙ n dS) = ∫∫[D] ((R(x,y,u₂) - R(x,y,u₁))dA), which is what we had written before, and so the initial 3 equations we have sought out to prove, well, we proved one of them, and we could do this exact same sequence to prove the others by changing their region types.

 

First off, Stokes's Theorem is terribly complicated to prove, so we will only be proving a special case of it at the end of this section.  What's going on, basically, is that this theorem is basically a higher-dimensional version of Green's Theorem.  Instead of a double integral over a region D related to a line integral around its boundary curve, we're going to relate a surface integral over a surface S to a line integral around the curve of S (which is a space curve).

If you have a surface S and you pick some curve along it and move counterclockwise, you are, ultimately, in positive orientation, so we're going to go with that.  Stokes's Theorem says

Let S be an oriented piece-wise smooth surface that is bounded by a simple, closed, piece-wise smooth boundary curve C with positive orientation.  Let F be a vector field whose components have continuous partial derivatives on an open region in R³ that contains S.  Then

∫[C]  (F ∙ dr) = ∫∫[S] (curl F ∙ dS)

Because ∫[C] (F ∙ dr) = ∫[C] (F ∙ T ds) and ∫∫[S] (curl F ∙ dS) = ∫∫[S] (curl F ∙ n dS), we say that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral of the normal component of the curl of F.  With a positive orientation, we usually write this as

∫∫[S] (curl F ∙ dS) = ∫[∂S] (F ∙ dr)

Interestingly, if the surface is flat and is on the XY plane with upward orientation (like a vertical cylinder), the unit normal is k, the surface integral becomes a double integral, and the theorem becomes

∫[C] (F ∙ dr) = ∫∫[S] (curl F ∙ dS) = ∫∫[S] ((curl F) ∙ k dA)

And that is precisely the vector form of Green's Theorem, so we can see that Green's Theorem is merely a special case of Stokes's Theorem.

So, let's do an example!

We wanna compute ∫∫[S] (curl F ∙ dS), where F(x,y,z) = xzi + yzj + xyk, and S is part of the sphere x² + y² + z² = 4 that lies inside the cylinder x² + y² = 1 and above the XY plane.

To find our C, we gotta solve those two S equations in there, which can EASILY be done by simply subtracting one from the other, and we get z² = 3, or z = sqrt(3).  Since this is a circular area, we can write a vector equation C as r(t) = cos(t)i + sin(t)j + sqrt(3)k.

F(r(t)) = sqrt(3)cos(t)i + sqrt(3)sin(t)j + cos(t)sin(t)k

r'(t) = -sin(t)i + cos(t)j

F ∙ dr = -sqrt(3)cos(t)sin(t) + sqrt(3)cos(t)sin(t) = 0

So, by Stokes's Theorem, we have ∫∫[S] (curl F ∙ dS) = ∫[C] (F ∙ dr) = ∫[0,2π] (0dt) = 0.

Lastly, if S₁ and S₂ are oriented surfaces with the same oriented boundary curve C and both satisfy the hypothesis of Stokes's Theorem, then ∫∫[S₁] (curl F ∙ dS) = ∫[C] (F ∙ dr) = ∫∫[S₂] (curl F ∙ dS).

That tells us, really, that sometimes it's difficult to integrate one way, but if we integrate over a different way, we'll get the same easier, so we have a means of getting a simpler manner, just like when we were computing multiple integrals and we chose one region over another.

So, let's work some problems!

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1)   Use Stokes's Theorem to evaluate ∫∫[S] (curl F ∙ dS), where F(x,y,z) = yzi + xzj + xyk, S is the part of the paraboloid z = 9 - x² - y² that lies above the plane z = 5, oriented upward.

If you plug in z = 5 to the paraboloid and rearrange, you find the intersection at x² + y² = 4, which is a circle of radius 2, so a vector equation is r(t) = 2cos(t)i + 2sin(t)j + 5k.

F(r(t)) = 10sin(t)i + 10cos(t)j + 4cos(t)sin(t)k

r'(t) = -2sin(t)i + 2cos(t)j

F ∙ dr = -20sin²(t) + 20cos²(t) = 20(cos²(t) - sin²(t)).

So, ∫∫[S] (curl F ∙ dS) = ∫[C] (F ∙ dr) = 20*∫[0,2π] (cos²(t) - sin²(t)  dt)

cos²(t) - sin²(t) = cos(2t) because sin²(t) = 1/2 - cos(2t)/2, cos²(t) = 1/2 + cos(2t)/2, and doing the algebra yields only cos(2t).  Integrating cos(2t) gives sin(2t)/2 from 0 to 2π, both are trivial, so our end result is just 0.

2)  Use Stokes's Theorem to evaluate ∫[C] (F ∙ dr), where F(x,y,z) = xi + yj + (x² + y²)k, and C is the boundary of the part of the paraboloid z = 1 - x² - y² in the first octant.

If on F, P = x, Q = y, and R = x² + y², then ∂R/∂y = 2y, ∂Q/∂z = 0, ∂P/∂z = 0, ∂R/∂x = 2x, ∂Q/∂x = 0, ∂P∂y = 0, so we have curl F = 2yi + 2xj.

If z = g(x,y) = 1 - x² - y², then ∂g/∂x = -2x, ∂g/∂y = -2y, and so we finally have ∫[C] (F ∙ dr) = ∫∫[S] (curl F ∙ dS) = ∫∫[D] (2y(-2x) - 2y(-2x)  dA) = ∫∫[D] (0dA) = 0.

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Why do you care?  The point of this is to shine some light on the curl vector, basically.  We can use it to prove that if curl F = 0 on all of R³, then F is conservative, but as for immediate physical applications, this is more of yet another way to understand the curl vector's vector field analyses.  It's just a better way to relate things to other things, really.  Also a note:  not every answer is 0.  XFD  I realize the example AND the two problems had answers = 0.  It was pure coincidence.

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Proof of special case:

Assume that the surface S has an equation z = g(x,y), (x,y) in D, where g has continuous second-order partial derivatives and D is a simple plane region whose boundary curve C₁ corresponds to C.  If the orientation of S is upward, then the positive orientation of C corresponds to the positive orientation of C₁.  We say that F = Pi + Qj + Rk, where the partial derivatives of P,Q,R are continuous.

Since S is now a graph of a function, we can

∫∫[S] (curl F ∙ dS) =
∫∫[D] ([(∂z/∂x)(∂R/∂y - ∂Q/∂z) - (∂z/∂y)(∂P/∂z - ∂R/∂x) + (∂Q/∂x - ∂P/∂y)]dA)

where the partial derivatives of P, Q, and R are evaluated at (x, y, g(x,y)).  If x = x(t), y = y(t), a < t < b is a parametric representation of C₁, then a parametric representation of C is x = x(t), y = y(t), z = g(x(t), y(t)), a < t < b.

And so, with Chain Rule, we can do

∫[C] (F ∙ dr) = Int[a,b] ((P(dx/dt) + Q(dy/dt) + R(dz/dt))dt)
= ∫[a,b] ((P(dx/dt) + Q(dy/dt) + R((∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)))dt)
= ∫[a,b] (((P + R(∂z/∂x))(dx/dt) + (Q + R(∂z/∂y))(dy/dt))dt)
= ∫[C₁] ((P + R(∂z/∂x))dx + (Q + R(∂z/∂y))dy)
= ∫∫[D] (((∂/∂x)(Q + R(∂z/∂y)) - (∂/∂y)(P + R(∂z/∂x)))dA) by Green's Theorem.

Chain Rule again, with P,Q,R as functions of x,y,z and z is a function of x,y gives

∫[C] (F ∙ dr) = ∫∫[D] (((∂Q/∂x) + (∂Q/∂z)(∂z/∂x) + (∂R/∂x)(∂z/∂y) + (∂R/∂z)(∂z/∂x)(delz/dely) + R(∂²z/(∂x∂y))) - ((∂P/∂y) + (∂P/∂z)(∂z/∂y) + (∂R/∂y)(∂z/∂x) + (∂R/∂z)(∂z/∂y)(∂z/∂x) + R(∂²z/(∂y∂x)))dA)

Four of those terms cancel, and the remaining can spit us what we saw with the first bit of IntInt[S] (curl F ∙ dS), and so we can conclude

∫[C] (F ∙ dr) = ∫∫[S] (curl F ∙ dS)

Well, this is it. This is the big section that almost all of calculus has ultimately been created from (well, not really. This is more of a "let's turn what we created into a general form, and then rejustify that what we created are 'special cases' of these forms."). The relationship we're looking for is the relationship between surface integrals and surface area, much as we saw a relationship between line integrals and arc length.

Suppose f is a function of three variables whose domain includes a surface S. We divide S into patches, S_ij with area ∆S_ij, and we evaluate f at a point P_ij* in each patch, multiply by the area ∆S_ij, and form the sum

∑[i=1:m]∑[j=1:n] (f(P_ij*)∆S_ij)

We take the limit as the patch size approaches 0, and so we define the surface integral of f over the surface S as

∫∫[S] (f(x,y,z)dS) = lim ∑[i=1:m]∑[j=1:n] (f(P_ij*)∆S_ij) as m,n -> ∞

And we can approximate the patch area ∆S_ij by the area ∆T_ij of an approximating parallelogram in the tangent plane, and so the limit becomes a double integral. We have two main types of surfaces: graphs and parametric surfaces. We'll start with graphs.

If the surface S is a graph of a function of two variables, then it has some equation z = g(x,y), (x,y) in D, and we assume the parameter domain D is a rectangle, and we divide this into smaller rectangles R_ij of equal size. The patch S_ij lies directly above the rectangle R_ij and the point P_ij* in S_ij is of the form (x_i*, y_j*, g(x_i*,y_j*)). We use the approximation

∆S_ij ~ ∆T_ij = sqrt([g_x(x_i,y_j)]² + [g_y(x_i,y_j)]² + 1)∆A which we learned in Chapter 16. If f is continuous on S and g has continuous partial derivatives, then we can ultimately say

∫∫[S] (f(x,y,z)dS) = lim ∑[i=1:m]∑[j=1:n] (f(x_i*,y_j*,g(x_i*,y_j*))sqrt([g_x(x_i,y_j)]² + [g_y(x_i,y_j)]² + 1)∆A) as m,n -> ∞ = ∫∫[D] (f(x,y,g(x,y))sqrt([g_x(x,y)]² + [g_y(x,y)]² + 1)dA)

And we usually write that, then, as this main formula:

∫∫[S] (f(x,y,z)dS) = ∫∫[D] (f(x,y,g(x,y))sqrt((∂z/∂x)² + (∂z/∂y)² + 1)dA)

And you can switch a few things around if you were wanting to project onto the XZ or YZ plane instead of the XY plane, creating a function h(x,z) or m(y,z) or something and then having f(m(y,z), y, z) or what have you.

Before we do an example, let's handle parametric surfaces. Suppose S has a vector equation r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k, (u,v) in D. We'll assume the parameter domain D is rectangular, divide it up into subrectangles R_ij with dimensions ddu and ddv, then the surface is divided into subpatches S_ij with areas ∆S_ij, and we recall from the previous section that

∆S_ij ~ |r_u x r_v|∆u∆v, where r_u = (∂x/∂u)i + (∂y/∂u)j + (∂z/∂u)k, r_v = (∂x/∂v)i + (∂y/∂v)j + (∂z/∂v)k are the tangent vectors at a corner of S_ij. If the components are continuous and r_u and r_v are nonzero and nonparallel in the interior of D, then we can show that even if D is not rectangular, we get

∫∫[S] (f(x,y,z)dS) = ∫∫[D] (f(r(u,v))|r_u x r_v|dA)

Also, if we regard our parametric surface as x = x, y = y, z = g(x,y), then |r_x x r_y| = sqrt(1 + (∂z/∂x)² + (∂z/∂y)²), which ultimately spits out the same formula as the one for graphs.

So, example time!

We wanna find the surface integral ∫∫[S] (x²dS), where S is the unit sphere x² + y² + z² = 1.

Well, we can parameterize this as x = sin(φ)cos(θ), y = sin(φ)sin(θ), z = cos(φ), 0 < φ < π, 0 < θ < 2π.

So, we have r(φ,theta) = sin(φ)cos(θ)i + sin(φ)sin(θ)j + cos(φ)k, and SKIPPING A BUNCH OF STEPS I DON'T WANT TO DO, |r_φ x r_θ| = sin(φ)

So, we have ∫∫[S] (x²dS) = ∫∫[D] ([sin(φ)cos(θ)]²|r_φ x r_θ|dA)
= ∫[0,2π]Int[0,π] (sin²(φ)cos²(θ)sin(φ)dφdθ)
= ∫[0,2π] (cos²(θ)dθ) * ∫[0,π] (sin³(φ)dφ)
= (1/2)*∫[0,2π] ((1 + cos(2θ))dθ) * ∫[0,π] ((sin(φ) - sin(φ)cos²(φ))dφ)

That first bit turns into θ + (1/2)sin(2θ) from 0 (trivial) to 2π gives 2π, then multiplying by (1/2) gives π.

The second bit turns into -cos(φ) + (1/3)cos³(φ) from 0 to π, which ultimately turns into 4/3. So, we have (4/3)*π = 4π/3.

Now, let's move on to oriented surfaces. Before we move on, we're going to remove certain types of surfaces. Namely, those that cannot be distinctly oriented, such as the Mobius strip, which has only one edge (take a strip of paper, twist one end 180 degrees, then tape the ends together. Tada. Traveling your finger along one side will have you loop around it until you cover both sides, basically, making it effectively "one sided.").

The surfaces we're gonna consider have at least two distinct sides. We'll start with some surface S that has a tangent plane at every point (x,y,z) on S (except at any boundary points). There are two unit normal vectors, n₁ and n₂ = -n₁ at (x,y,z). Basically, on, say, a flat sheet of paper, one points up and the other points down at the same point. If it is possible to choose some unit normal vector n at every point (x,y,z) so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides S with an orientation. There are two possible orientations. With our flat sheet of paper, basically, ALL n point up, or ALL n point down, and either will be our orientation.

For a surface z = g(x,y) given as a graph of g, the induced orientation is given by the unit normal vector

n = ((-∂g/∂x)i - (∂g/∂y)j + k)/(sqrt(1 + (∂g/∂x)² + (∂g/∂y)²).

Because the k component is positive, we have an upward orientation of the surface.

If S is a smooth orientable surface given in parametric form by a vector function r(u,v), then it is automatically supplied with the orientation of the unit normal vector

n = (r_u x r_v) / |r_u x r_v|

And the opposite orientation is given by -n.

For a closed surface, the GENERAL convention (pretty damn standard around the world) is that the positive orientation is one in which ALL unit normal vectors point outward. A negative orientation is one in which all unit normal vectors point inward the surface.

So, since we BROUGHT UP vectors, we're probably interested in, say, surface integrals of vector fields! We are!

Suppose that S is an oriented surface with a unit normal vector n, and we imagine a fluid with density ρ(x,y,z) and velocity field v(x,y,z) flowing through S, and S does not impede the fluid's flow (like a fish net in a stream).  Then the rate of flow per unit area is ρv.  If we divide S into small patches S_ij, then S_ij is nearly planar, and so we can approximate the mass of the fluid crossing S_ij in the direction of the normal n per unit time by the quantity

(ρv ∙ n)A(S_ij), where ρ, v, n are evaluated at some point on S_ij, and A(S_ij) is simply the area of S_ij.  By summing all these quantities and taking the limit, we get a surface integral of the function ρv ∙ n over S, or

∫∫[S] (ρv ∙ n dS) = ∫∫[S] (ρ(x,y,z)v(x,y,z) ∙ n(x,y,z) dS).

Now, if we write F = ρv, then F is also a vector field on R³, and the integral becomes

∫∫[S] (F ∙ n dS).  And now we can get away from our example and create a form for any generic vector field:

If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS), also known as the flux of F across S.

That's a general form.  What we want to do is look at our two types of surfaces again:  graphs and parametrics.  We'll start with graphs again.

S is z = g(x,y), and we can find n by noting that S is also the level surface f(x,y,z) = z - g(x,y) = 0, and the gradient Ñf(x,y,z) is normal to this surface at (x,y,z), so a unit normal vector is

n = Ñf(x,y,z) / |Ñf(x,y,z)|
= (-g_x(x,y)i - g_y(x,y)j + k) / sqrt(1 + [g_x(x,y)]² + [g_y(x,y)]²)

Since k is positive, this is the upward unit normal.  We then want to say F(x,y,z) = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k, and so

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS)

= ∫∫[D] ((Pi + Qj + Rk) ∙ [((-∂g/∂x)i + (-∂g/∂y)j + k)/(sqrt(1 + (∂g/∂x)² + (∂g/∂y)²))]sqrt((∂g/∂x)² + (∂g/∂y)² + 1)dA)

And so, we can rewrite all that in a MUCH simpler form:

∫∫[S] (F ∙ dS) = ∫∫[D] ((-P(∂g/∂x) - Q(∂g/∂y) + R)dA)

For a downward unit normal force, we just multiply everything by -1.  And, of course, we don't have to have z = g(x,y), we can have y = h(x,z) or x = m(y,z) and replace things appropriately.

Now let's look at the parametric surfaces!

If S is given by a vector function r(u,v), then n is given by our cross product mess we saw earlier, and so

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ ([(r_u x r_v)/|r_u x r_v|]dS)
= ∫∫[D] ((F(r(u,v)) ∙ [(r_u x r_v)/|r_u x r_v|])*|r_u x r_v|dA), where D is the parameter domain, and so we can simplify that greatly to

∫∫[S] (F ∙ dS) = ∫∫[D] (F ∙ (r_u x r_v)dA)

And we can make a quick note that r_x x r_y = (-∂g/∂x)i - (∂g/∂y)j + k from a previous analysis a few paragraphs up, so we can turn the parametric surface into the graphical equation.

And there are many applications to these!  Suppose E is an electric field, then the surface integral ∫∫[S] (E ∙ dS) is called the electric flux of E through the surface S.  There is also an important law in electrostatics called Gauss's Law, which says that the net charge enclosed by a closed surface S is Q = ε₀*∫∫[S] (E ∙ dS), where ε₀ is a constant called the permittivity of free space.

Also, if the temperature at a point (x,y,z) is u(x,y,z) in a body, then the heat flow is defined as the vector field  F = -KÑu, where K is an experimentally determined constant called the conductivity of the substance.  The rate of heat flow across the surface S in the body is then given by the integral  ∫∫[S] (F ∙ dS) = -K*∫∫[S] (Ñu ∙ dS).

So, let's do an example!

The temperature u in a metal ball is proportional to the square of the distance from the center of the ball.  Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.

The center of the ball will be origin, so we have u(x,y,z) = C(x² + y² + z²), where C is some proportionality constant as per our problem.  Then the heat flow F(x,y,z) = -KÑu = -KC(2xi + 2yj + 2zk), where K is the conductivity of the metal.

We can find that the outward unit normal to the sphere x² + y² + z² = a² at (x,y,z) is

n = (1/a)(xi + yj + zk), so

F ∙ n = (-2KC/a)(x² + y² + z²), but that last bit is also our S, so we can replace that with a², and then our multiplication simplifies to just -2aKC.  Thus,

∫∫[S] (F ∙ dS) = ∫∫[S] (F ∙ n dS) = -2aKC*∫∫[S] (dS)
= -2aKCA(S) = -2aKC(4πa²) = -8KCπa³.

Tada.  Now for some more problems!  XD

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1)  Evaluate ∫∫[S] (ydS) where S is the surface z = (2/3)(x^3/2 + y^3/2), 0 < x < 1, 0 < y < 1.

Well, we need ∂z/∂x and ∂z/∂y:  ∂z/∂x = sqrt(x), ∂z/∂y = sqrt(y), so we have

∫∫[D] (ysqrt(sqrt(x)² + sqrt(y)² + 1)dA) = ∫[0,1]∫[0,1] (ysqrt(x + y + 1)dxdy).

If we let A = y + 1 (because we're doing with x first, so y is a pretend constant), we have sqrt(x + A), and we say u = x + A, du = dx, so we have sqrt(u), and integrating that gives (2/3)u^3/2, or (2/3)(x + y + 1)^3/2 from x = 0 to 1, giving us (2/3)[(2 + y)^3/2 - (1 + y)^3/2].  So now we just have (2/3)*∫[0,1] ([(2 + y)^3/2 - (y + 1)^3/2]dy).

We break them, and for the first, we say u = 2 + y, so du = dy, and we have (u -2)u^3/2, since y = u - 2, which is u^5/2 - 2u^3/2, which integrates to (2/7)u^7/2 - (4/5)u^5/2 from u = 2 to u = 3 (by plugging in our boundaries for y = 0 and y = 1 into u = 2 + y), so we have [(2/7)3^7/2 - (4/5)3^5/2] - [(2/7)2^7/2 - (4/5)2^5/2].

For the second, we say u = y + 1, so y = u - 1, and we have (u - 1)u^3/2, or u^5/2 - u^3/2, which integrates to (2/7)u^7/2 - (2/5)u^5/2 from u = 1 to u = 2, so we have [(2/7)2^7/2 - (2/5)2^5/2] - [(2/7) - (2/5)]

Putting it all together, we have (2/3)*[[(2/7)3^7/2 - (4/5)3^5/2 - (2/7)2^7/2 + (4/5)2^5/2] - [(2/7)2^7/2 - (2/5)2^5/2 + 4/35]] = (2/3)(18sqrt(3)/35 + 8sqrt(2)/35 - 4/35) = (4/105)(9sqrt(3) + 4sqrt(2) - 2).

2)  Evaluate ∫∫[S] (F ∙ dS) for F(x,y,z) = xyi + 4x²j + yzk, S is the surface z = xe^y, 0 < x < 1, 0 < y < 1, upward orientation.

We can say P = xy, Q = 4x², and R = yz, then z = g(x,y) = xe^y.  So, ∂g/∂x = e^y, ∂g/∂y = xe^y, and so -P(∂g/∂x) = -xye^y, -Q(∂g/∂y) = -4x³e^y, and we have

∫∫[S] (F ∙ dS) = ∫∫[D] ((-P(∂g/∂x) - Q(∂g/∂y) + R)dA)
= ∫[0,1]∫[0,1] ((-xye^y - 4x³e^y + yxe^y)dydx)

The first and last terms cancel, leaving us only with -4x³e^ydydx, and integrating that with respect to y gives us just -4x³e^y from y = 0 to 1 gives us (-4x³e + 4x³) = 4(1 - e)x³.  So, we have 4(e + 1)*∫[0,1] (x³dx), which spits out an x⁴/4 from 0 (trivial) to 1 gives us 1/4, and so we have (1/4)*4*(1 - e) = 1 - e

3)  A fluid has density 1500 and velocity field v = -yi + xj + 2zk; find the rate of flow outward through the sphere x² + y² + z² = 25.

F = ρv = -1500yi + 1500xj + 3000zk.  We see the radius = 5, and if we convert to spherical coordinates, we have x = 5sin(φ)cos(θ), y = 5sin(φ)sin(θ), z = 5cos(φ), and we can generate an r(φ,θ).

r_φ x r_θ = det([[i  j  k][∂x/∂φ  ∂y/∂φ  ∂z/∂φ][∂x/∂θ  ∂y/∂θ  ∂z/∂θ]]) = det([[i  j  k][5cos(φ)cos(θ)  5cos(φ)cos(θ)  -5sin(φ)][-5sin(φ)sin(θ)  5sin(φ)cos(θ)  0]]), you can do that mess yourself because I am tired of doing these damned Greek letters, and you get

25sin²(φ)cos(θ)i + 25sin²(φ)sin(θ)j + 25sin(φ)cos(φ)k.  Our boundaries are 0 < φ < π, 0 < θ < 2π, and so our double integral is just

F ∙ (r_φ x r_θ) = -1500*5sin(φ)sin(θ)*25sin²(φ)cos(θ) + 1500*5sin(φ)cos(θ)*25sin²(φ)cos(φ) + 3000*5cos(φ)*25sin(φ)cos(φ)
= 1500*[-125sin³(φ)sin(θ)cos(θ) + 125sin³(φ)sin(θ)cos(θ) + 250sin(φ)cos²(φ)]

∫∫[S] (F ∙ dS) = 1500*∫[0,2π]∫[0,π] ((-125sin³(φ)sin(θ)cos(θ) + 125sin³(φ)sin(θ)cos(θ) + 250sin(φ)cos²(φ) )dφdθ).  A VERY quick analysis shows that the first two terms are GONE, leaving us with only

∫[0,2π]∫[0,π] (250sin(φ)cos²(φ)dφdθ).  Let u = cos(φ), du = -sin(φ)dφ, so we have -u², which is just -u³/3 or -cos³(φ)/3 from 0 to π, or cos³(0)/3 - cos³(π)/3 = 1/3 + 1/3 = 2/3.  Integrating that to θ, we get 2θ/3 from 0 (trivial) to 2π gives us 4π/3.  We multiply that by 250 and 1500, and we have a final answer of 500000π.

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Why do you care?  Surface integrals are pretty much everything, as everything can be simplified from surface integrals to some degree.  When you're studying fluid mechanics, you have to measure things called control surfaces and control volumes, both of which are surface integrals.  When studying heat, you do the same thing (well, heat IS a fluid).  Studying electricity and magnetism, you do the same.  Studying stress/shear in materials for structures, you're looking at more surface integrals.  Surface integrals are everywhere, and they're the only ways we can even have our physics.

 

Pretty big section with five topics to cover, and they all go in a nice order.  Way back in Chapter 9, we found the area of a surface of revolution.  In Chapter 16, we found the area of a surface with an equation z = f(x,y).  Now we're interested in any general surface.

When we describe a space curve, we do so usually with some r(t) = x(t)i + y(t)j + z(t)k.  However, that was just ONE parameter, so it was a curve.  If we introduce a second parameter, in much how z = f(x,y) gives us a surface, we get a, uh, a surface.  XD  We suppose that

r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k is a vector-valued function defined on a region D in the UV-plane.  So, x,y,z are the component functions of r, and they are functions of two variables with the domain D.  We can classify the set of all points (x,y,z) in R³ as

x = x(u,v), y = y(u,v), z = z(u,v), and (u,v) varies throughout D, and we all these sets of points a parametric surface S.  Obviously, those three equations are parametric equations.  If r is like a piece of string with one end tied to the origin, as we move the string along (as u and v change), the tip of the free end of the string creates a surface, and this is our surface.

Most computer algebraic systems, such as Mathematica and the like, generate something called grid curves, which you've seen as the surface with a bunch of lines through it.  What's being done is that the software, which we can also do without software though drawing 3-D can be tricky, analyzes two families of curves found on S.  One is a family where u remains constant, so only v varies.  The other is the opposite.  By keeping u constant, we say u = u₀, and so our parametric surface vector function is just r(u₀,v), which is a function of only ONE variable, and so it is only a curve, C₁.  Doing the same for v, we have r(u,v₀), and this creates a curve C₂, and then we simply "overlaps" the two curves to create what you typically see in software.

Now, the trick is, then, finding a parametric representation of a surface.  Let's start with an easy example.

We wanna find the parametric representation of the sphere x² + y² + z² = a².

Well, in spherical coordinates, we can simply say ρ = a, since ρ is a radius and the radius of that sphere is just a.  So, we can choose some angles φ and θ in spherical coordinates as the parameters, and we can EASILY get the following parametric equations:

x = asin(φ)cos(θ), y = asin(φ)sin(θ), z = acos(φ)

The corresponding vector equation, then, is r(φ,θ) = asin(φ)cos(θ)i + asin(φ)sin(θ)j + acos(φ)k.

r is only a function of φ and theta because our radius was a constant at a.  Makes sense.  A sphere doesn't have variable radius.  And we also say 0 < φ < π, 0 < θ < 2π, so our parameter domain is the rectangle D = [0,π] x [0,2π].

Ahh, but we already knew spherical coordinates from a previous section, so that was too easy.  So, let's try a different one.

Let's parameterize and find the vector function of the elliptic paraboloid z = x² + 2y².

The simplest way of tackling any surface is to do the most trivial case.  That is, we'll say x = x, y = y, and then z = x² + 2y².  Our vector function, then, is r(x,y) = xi + yj + (x² + 2y²)k.

And this sort of thinking makes sense.  In the previous chapters, we considered x its own independent variable, y its own independent variable, and z = f(x,y).  So, doing THAT can make doing THIS rather trivial, for we'll always have some r(x,y) = xi + yj + f(x,y)k.

Let's do one last example to show that, but also show something else.

We wanna parameterize z = 2sqrt(x² + y²) and find the vector function.

x = x, y = y, z = 2sqrt(x² + y²), so r(x,y) = xi + yj + 2sqrt(x² + y²)k.

BUT, that is only ONE possibility!  We could have converted to, say, cylindrical coordinates, where x = rcos(θ), y = rsin(θ), and z = z.  If x² + y² = r², then sqrt(r²) = r since a radius can't be negative, and so 2sqrt(x² + y²) = 2*r = 2r.  So, we could INSTEAD have THESE parametrics and vector function:

x = rcos(θ), y = rsin(θ), z = 2r, r(r,θ) = rcos(θ)i + rsin(θ)j + 2rk, where r > 0 and 0 < θ < 2π.

And that exemplifies our last point for this topic:  there is always more than one way to parameterize a surface.  You just gotta go with what seems most convenient, or I guess for your basic calculus class, go with whatever gets through the problem fastest and still correct.

Our next topic has us dealing with surfaces of revolutions.  If we have some curve y = f(x), a < x < b, and we rotate y about the X-axis, we can generate a surface S.  If we let theta be the angle of actual rotation (how far we wanna rotate it, which we may not WANT to completely rotate it around a full circle like we did in Chapter 9), then if (x,y,z) is a point on S, we can parameterize S as

x = x, y = f(x)cos(θ), z = f(x)sin(θ).  And so, we have a < x < b, 0 < θ < 2π as our domain.

A quick example.  Let's parameterize the surface generated by rotating y = 1/x, 1 < x < 3, halfway about the X-axis.

We get x = x, y = (1/x)cos(θ), z = (1/x)sin(θ), 1 < x < 3, and 0 < θ < π (halfway of a full revolution is just half a circle, which is π).  And then we can say r(x,θ) = xi + (1/x)cos(θ)j + (1/x)sin(θ)k.

Next topic has us dealing with tangent planes!

We go back to our thing about grid curves and holding one variable constant.  We want to find the tangent plane to a vector r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k at some point P₀ with a position vector r(u₀,v₀).  Keeping the same notation, r(u₀,v) will generate C₁ and r(u,v₀) will generate C₂.  The tangent vector to C₁, then, is simply taking the partial derivative of r with respect to v, as it is the only variable, so we get r_v = (∂x/∂v)(u₀,v₀)i + (∂y/∂v)(u₀,v₀)j + (∂z/∂v)(u₀,v₀)k, and we can do the SAME analysis for r_u.

If r_u x r_v is not 0, then we call S smooth (has no corners).  For a smooth surface, the tangent plane is the plane that contains the tangent vectors r_u and r_v, and the vector r_u x r_v is a normal vector to the tangent plane.

So, a quick example.  We wanna find the tangent plane to the surface with parametrics x = u², y = v², z = u + 2v at the point (1,1,3).

We should know how to find partial derivatives by now, so I'll go ahead and say r_u = 2ui + k, r_v = 2vj + 2k.

The cross product is det([[i  j  k][2u  0  1][0  2v  2]]) = -2vi - 4uj + 4uvk.  The point (1,1,3) corresponds to simply u = 1, v = 1 (and if you do x = u² = 1, y = v² = 1, z = 1 + 2(1) = 3, so it matches up!), so we plug in those to our normal vector to get -2i - 4j + 4k.

Thus, our equation of our tangent plane is -2(x - 1) - 4(y - 1) + 4(z - 3) = 0, and we can simplify that to x + 2y - 2z + 3 = 0.

Slowly building up.  Now we can identify surfaces and their tangent planes regardless of how they're created.  Now we are interested in surface area, our next topic!

For simplicity, let's start by looking at a surface whose domain D is a rectangle, and we divide it into subrectangles R_ij.  We choose some (u_i*, v_j*) to be the lower left corner of R_ij.  The part S_ij of the surface S that corresponds to R_ij is called a patch, and has the point P_ij with the position vector r(u_i*,v_j*) as one of its corners.  Let r_u* = r_u(u_i*,v_j*) and r_v* = r_v(u_i*,v_j*) be the tangent vectors at P_ij as given by our earlier definition a few paragraphs above in the previous topic.

The two edges of our patch that meet at P_ij can be approximated by vectors.  These vectors can be approximated by the vectors ∆ur_u* and ∆vr_v* because partial derivatives can be approximated by difference quotients.  So, we approximate S_ij by a parallelogram determined by these two vectors, and this parallelogram lies in the tangent plane to S at P_ij.  The area of this parallelogram is |(∆ur_u* x ∆vr_v*)| = |r_u* x r_v*|∆u∆v, and so to approximate the area of S we take the double sum, and we can turn that into a Riemann sum and then a double integral ∫∫[D] (|r_u x r_v|dudv).  Thus, we get the following definition:

If a smooth parametric surface S is given by the equation r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k, (u,v) in D, and S is covered just once as (u,v) ranges throughout the parameter domain D, then the surface area of S is

A(S) = ∫∫[D] (|r_u x r_v|dA), where r_u = (∂x/∂u)i + (∂y/∂u)j + (∂z/∂u)k, r_v = (∂x/∂v)i + (∂y/∂v)j + (∂z/∂v)k.

Phew.  Example time!

We wanna find the surface of a sphere with radius r.

Well, x = rsin(φ)cos(θ), y = rsin(φ)sin(θ), z = rcos(φ), D = {(φ,θ) | 0 < φ < π, 0 < θ < 2π}

First, we need the cross product of the two tangent vectors:

r_φ x r_θ = det([[i  j  k][∂x/∂φ  ∂y/∂φ  ∂z/∂φ][∂x/∂θ  ∂y/∂θ  ∂z/∂θ]]) = det([[i  j  k][rcos(φ)cos(θ)  rcos(φ)sin(θ)  -rsin(φ)][-rsin(φ)sin(θ)  rsin(φ)cos(θ)  0]])
= r²sin²(φ)cos(θ)i + ^r2sin²(φ)sin(θ)j + r²sin(φ)cos(φ)k.

|r_φ x r_θ| = sqrt(r⁴sin⁴(φ)cos²(θ) + r⁴sin⁴(φ)sin²(θ) + r⁴sin²(φ)cos²(φ)), and you can see the first two terms turn just into r⁴sin⁴(φ) because of the trig identity in there, so we have

= sqrt(r⁴sin⁴(φ) + r⁴sin²(φ)cos²(φ)).  The r⁴ comes out to r² outside the radical, and we can say sin⁴(x) = sin²(x)sin²(x), so we get r⁴sqrt(sin²(φ)) = r²sin(φ).

No absolute values because of our domain always being positive.  Thus, the area is just

A = ∫∫[D] (|r_φ x r_θ|dA) = ∫[0,2π]∫[0,π] (r²sin(φ)dφdθ) = r²*∫[0,2pi] (dθ) *∫[0,pi] (sin(φ)dφ).

The first integral is just θ from 0 (trivial) to 2π, so it's r²2pi*stuff.  The second bit is -cos(φ) from 0 to π, so we have -cos(π) + cos(0) = 1 + 1 = 2, so we have r²*2π*2 = 4πr².  Tada!

Now for our LAST topic!

Now we're interested in the surface area of a graph of a function!  That is, we've a special case of a surface S with an equation z = f(x,y), where (x,y) lies in D and f has continuous partial derivatives, and x and y are parameters, so we have

x = x, y = y, z = f(x,y).  Then r_x = i + (∂f/∂x)k, r_y = j + (∂f/∂y)k, and so

r_x x r_y = det([[i  j  k][1  0  ∂f/∂x][0  1  ∂f/∂y]]) = -(∂f/∂x)i - (∂f/∂y)j + k

|r_x x r_y| = sqrt((-∂f/∂x)² + (-∂f/∂y)² + 1) = sqrt(1 + (∂z/∂x)² + (∂z/∂y)²), and so we have our surface area formula:

A(S) = ∫∫[D] (sqrt(1 + (∂z/∂x)² + (∂z/∂y)²)dA) = ∫∫[D] (|r_x x r_y|dA), and the first formula here is what we learned in Chapter 16!  However, in Chapter 9, for a surface of revolution, we had something a little different.  That is, we said A(S) = 2pi*Int[a,b] (f(x)*sqrt(1 + [f'(x)]²)dx).  Are we still consistent?  YES!  ...But we'll show why at the very end!

Let's do an example instead!

We wanna find the area of the part of the paraboloid z = x² + y² that lies under the plane z = 9.

∂z/∂x = 2x, ∂z/∂y = 2y, so our integrand becomes sqrt(1 + (2x)² + (2y)²) = sqrt(1 + 4x² + 4y²) = sqrt(1 + 4(x² + y²)).  At the intersection of the plane z = 9, we have a disk centered at the origin with radius = 3, so we can simply say 0 < r < 3, 0 < θ < 2π, and we have cylindrical coordinates:

∫∫[D] (sqrt(1 + 4(x² + y²))dA) = ∫[0,2π]∫[0,3] (sqrt(1 + 4r²)rdrdθ)
= ∫[0,2π] (dθ) *∫[0,3] (rsqrt(1 + 4r²)dr)

The first bit is trivial and just becomes 2π.  The second bit we have u = 1 + 4r², so du = 8rdr, so we have (1/8)*∫(u^1/2du), which transforms into (1/8)*u^3/2/(3/2) = (2/3)*(1/8)u^3/2 or (1/12)(1 + 4r²)^3/2 from 0 to 3, so we have

(1/12)[(1 + 4*9)^3/2 - (1 + 0)^3/2] = (1/12)(37^3/2 - 1), and we multiply that by 2π to get

(π/6)(37^3/2 - 1).  Bam.  We're done, let's do some problems!

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1)  Find an equation of the tangent plane to the given parametric surface at the specific point:  r(u,v) = uvi + usin(v)j + vcos(u)k, u = 0, v = π.

r_u = vi + sin(v)j - vsin(u)k
r_v = ui + ucos(v)j + cos(u)k
r_u x r_v = det([[i  j  k][v  sin(v)  -vsin(u)][u  ucos(v)  cos(u)]])
= i(sin(v)cos(u) + uvcos(v)sin(u)) - j(vcos(u) + vusin(u)) + k(uvcos(v) - usin(v)).

We now just plug in 0 and π for u and v respectively, and we get
i(sin(π)cos(0) + 0*πcos(π)sin(0)) - j(πcos(0) + π*0sin(0)) + k(0*πcos(v) - 0*sin(π)) = -πj

To find our X,Y terms, we plug in for the original r(u,v) = r(0,π) = 0*πi + 0*sin(π)j + π*cos(0)k = πk, or (0,0,pi).  So our tangent plane equation, then, is

0(x - 0) + -π(y - 0) + 0(z - π) = 0, or -πy = 0, or y = 0.

2)  Find the area of the part of the surface z = 1 + 3x + 2y² that lies above the triangle with vertices (0,0),(0,1),(2,1).

x = x, y = y, z = 1 + 3x + 2y², r(x,y) = xi + yj + (1 + 3x + 2y²)k, D = {(x,y) | 0 < x < 2, 0 < y < x/2}
r_x = i + 3k
r_y = j + 4yk
r_x x r_y = det([[i  j  k][1  0  3][0  1  4y]])
= (0 - 3)i - (4y - 0)j + (1 - 0)k = -3i - 4yj + k.
|r_x x r_y| = sqrt((-3)² + (-4y)² + 1²) = sqrt(9 + 1 + 16y²) = sqrt(10 + 16y²).  So, we have

A(S) = ∫∫[D] (sqrt(10 + 16y²)dA) = ∫[0,2]∫[0,x/2] (sqrt(10 + 16y²)dydx).

But ew.  We can't really handle that with respect to y, so we'll switch our domain to be a different type region, a type 2 region.  Thus, D = {(x,y) | 0 < y < 1, 0 < x < 2y}.  Thus, we NOW have

∫[0,1]∫[0,2y] (sqrt(10 + 16y²)dxdy).  Integrating to x gives xsqrt(10 + 16y²) from 0 (trivial) to 2y gives

2ysqrt(10 + 16y²).  NOW we can handle that.  We say u = 10 + 16y², so du = 32ydy, and so we have (1/16)*Int(u^1/2du) which turns into (1/16)*(2/3)u^3/2, or (1/24)(10 + 16y²)^3/2 from 0 to 1, giving us

(1/24)[(10 + 16)^3/2 - (10 + 0)^3/2) = (1/24)[26^3/2 - 10^3/2].

3)  Find a parametric representation of the surface for the part of the plane z = x + 3 that lies inside the cylinder x² + y² = 1.

Well, right at where they intersect, looking down from above, we see only an ellipse on the plane z = x + 3.  So, if we parameterize into cylindrical coordinates, we have x = rcos(θ), y = rsin(θ)z = rcos(θ) + 3.  A vector form of this would be r(r,θ) = rcos(θ)i + rsin(θ)j + (rcos(θ) + 3)k, and then for BOTH, our domain is 0 < θ <π, 0 < r < 1.  This was made easy because rather than looking at it from strictly top, bottom, left, right, front, or back, we looked at it from the very plane we were analyzing.  You can look at things diagonally, can't you?

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Why do you care?  There's nothing terribly significant about what we learned here.  Rather, we are preparing for the next section which is easily one of the most important sections in all of calculus that ultimately unifies everything you've learned (which this section helped to do greatly) from "function" to "vector" notation.  As you have seen in this chapter, EVERYTHING you've done is simply a "particular case" of a general vector form, which this chapter is all about.  Technically, you've learned nothing new.  You've just learned a new way to approach it.  And without the fundamentals of this section, the next section frankly would not exist.

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Consistency in area of surface of revolution:

We say the surface S is obtained by rotating the curve y = f(x), a < x < b about the X-axis, where f(x) > 0 and f' is continuous.  Then our parametrics are x = x, y = f(x)cos(θ), z = f(x)sin(θ), a < x < b, 0 < θ < 2π.

r_x = i + f'(x)cos(θ)j + f'(x)sin(θ)k

r_θ = -f(x)sin(θ)j + f(x)cos(θ)k

r_x x r_θ = det([[i  j  k][1  f'(x)cos(θ)  f'(x)sin(θ)][0  -f(x)sin(θ)  f(x)cos(θ)]])
= f(x)f'(x)i - f(x)cos(θ)j - f(x)sin(θ)k

|r_x x r_θ| = sqrt([f(x)]²[f'(x)]² + [f(x)]²cos²(θ) + [f(x)]²sin²(θ))
= sqrt([f(x)]²[1 + (f'(x))²])
= f(x)sqrt(1 + [f'(x)]²)  since f(x) > 0

A = ∫∫[D] (|r_x x r_θ|dA)
= ∫[0,2π]∫[a,b] (f(x)sqrt(1 + [f'(x)]²)dxdθ)
= ∫[0,2π] (dθ) * ∫[a,b] (f(x)sqrt(1 + [f'(x)]²)dx)

The first bit just turns into θ, from 0 (trivial) to 2π, which is just

2π*∫[a,b] (f(x)sqrt(1 + [f'(x)]²)dx)