Question

Let F be a vector field F = (3x)i + (xz^3)j - (e^y)k.
Let B be a solid ball x^2 + y^2 + z^2 <= 1.
If n is the outward facing unit normal to S, use the divergence theorem to calculate the surface integral

Answer 1

I know that you can convert it to the triple integral of del F dV.

Answer 2

We have a solid \(B\) bounded by a surface \(S\).

Answer 3

We will let \(\mathbf S\) parametrize \(S\).\[
\iint_S\mathbf F \times d\mathbf S =\iint_S\mathbf F\cdot \mathbf n~dS= \iiint_V\nabla \cdot \mathbf F~dV
\]

Answer 4

So we want to gets \(B\)'s limits of integration: \[\begin{array}{rcl}
-\sqrt{1-(y^2+x^2)}<&z&<\sqrt{1-(y^2+x^2)}\\
-\sqrt{1-x^2}<&y&<\sqrt{1-x^2} \\
-1 < &x& < 1
\end{array}
\]

Answer 5

Then \(dV = dzdydx\).

Answer 6

\[
\nabla \cdot F = 3 + 0 - 0 = 3
\]

Answer 7

The only think we can do to make this any easier is to use spherical coordinates.

Answer 8

Gotcha. I didn't quite understand the limits, but I am familiar with how you set it up.

Answer 9

I would convert to spherical coords. I don't remember exactly certain details.

Answer 10

I believe the Jacobian for it is \(dz~dy~dz = \rho \sin \phi ~d\rho ~d\phi ~d\theta \)

Answer 11

rho^2 I believe

Answer 12

Well, who cares anyway: \[
\iiint_V 3~dV = 3 \iiint dV = 3V = 3\frac 43\pi (1)^3
\]

Answer 13

That was my other question. In all instances, if given that the object is of a known volume, you can substitute the triple integral of dV for the known formula, correct?

Answer 14

And since we calculate out del F to be 3, its just substitute in for r and multiply by 3, as you did.

Answer 15

No, you are assuming that \(\nabla \cdot \mathbf F\) is constant.

Answer 16

Only if del F is a constant*

Answer 17

Otherwise, you would integrate over the limits of \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{1} delop F \rho^2 \sin \phi d \rho d \phi d \theta \]

Answer 18

Given that this is a sphere.

Answer 19

yeah, I guess. So \(0<\phi <\pi\) then? I forgot.

Answer 20

For a sphere, yes.

Answer 21

Sweet, thanks again

Answer 22

If you happen to be bored, I'm asking one last question that concerns Stokes theorem. I've tried it a couple of times and I'm not sure where I'm going wrong exactly.

Answer 23

ok