mathThis 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 + exz^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 + exz^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 Ba is a ball with center P0 and a VERY small radius a, then div F(P) ~ div F(P0) for all points in Ba since div F is continuous. We approximate the flux over the boundary sphere Sa as
∫∫[Sa] (F ∙ dS) = ∫∫∫[Ba] (div F dV)
~ ∫∫∫[Ba] (div F(P0) dV)
= div F(P0) *V(Ba), where V(Ba) 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(Ba))*∫∫[Sa] (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] (Dnf dS) = ∫∫∫[E] (Ñ2f 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. Dnf = Ñf ∙ n, so
∫∫[S] (Dnf dS) = ∫∫[S] (Ñf ∙ n dS) = ∫∫∫[E] (div (Ñf)dV) = ∫∫∫[E] (Ñ2f 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.
