Question

[Calculus 3] If anyone can help me with the following integral, that'd be great! Thanks!

Answer 1

Use the Divergence Theorem to evaluate \[\int\limits_{ }^{ }\int\limits_{S}^{ }FdS\]where \[F = <3x^3,3y^3,z^3>\]and S is the sphere (x^2) + (y^2) + (z^2) = 1 oriented by the upward normal.

Answer 2

@mathmate might be able to help!

Answer 3

what does divergence thm say ?

Answer 4

Divergence Theorem states that I can convert a surface integral of a vector field F with respect to dS into a triple integral of the divergence of the vector field div(F) with respect to dV. I know my div(F) is right and I try to convert the bounds of the volumetric integral appropriately, but I cannot seem to get the final answer correct.

Answer 5

.

Answer 6

are you using spherical coordinates ?

Answer 7

At the very least, calculate the divergence. :P

Answer 8

I am converting to spherical coordinates. My bounds end up being rho going from 0 to 1, phi going from 0 to pi, and theta going from 0 to 2*pi. That's when I substitute appropriate spherical expressions for my div(F) and factor in the Jacobian.

Answer 9

Hint:

\[6z^2-6z^2=0\]

You can add 0 to anything and it doesn't change it, but sometimes it can make a big difference. ;P

Answer 10

My div(F) is 9(x^2) + 9(y^2) + 3(z^2), and after converting to spherical coordinates and simplifying appropriately (including factoring in the spherical Jacobian), I get my integral and integrand to be as follows: \[3 \int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi}\int\limits_{0}^{1}\rho^4(3 \sin^2 (\theta) \sin(\phi) + \sin(\phi) \cos^2 (\phi)) d \rho d \phi d \theta\]Am I setting this up wrong?

Answer 11

\[\begin{align}\iint\limits_{S}\vec{F}\cdot d\vec{S} &= \iiint\limits_D \text{div}(\vec{F})\, dV\\~\\
&= \iiint\limits_D 9x^2+9y^2+3z^2-6z^2\, dV\\~\\
&= \iiint\limits_D 9(x^2+y^2+z^2)-6z^2\, dV\\~\\
&= \iiint\limits_D 9\rho^2-6(\rho\cos\phi)^2\, dV\\~\\
\end{align}\]

Answer 12

Is that technique of adding and subtracting a common expression to net a simpler integrand a common one? It's useful, obviously, but I've never really seen it used yet, at least by my professor(s).

Answer 13

translating to spherical and simplifying the trig should give you the same expression

Answer 14

I attempted to translate to spherical and substitute any and all trig functions that could make it easier to integrate, but apparently I must've messed something up. Now that I'm working through it, I can tell immediately a few terms that I must've added unnecessarily in my initial integration attempts. Let's see if this time works!

Answer 15

setting up the integral, bounds, jacobian is the key. evaluating is not so much important, you may use wolfram to evaluate it

Answer 16

that 6z^2-6z^2 thingy is just an observation, final answer wont change if that idea doesn't strike to you

Answer 17

My final answer turned out to be 28*pi/5, which happens to be correct. Thanks for all your help, everyone! That's definitely an effective trick that I wasn't aware of, that method of adding/subtracting 'null' terms to further simplify an integrand expression. I'll definitely keep that in mind.

Answer 18

http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=89c969c21b169fa996f899d9b2a98588&title=Spherical%20Integral%20Calculator&theme=blue&i0=9%5Crho%5E2-6(%5Crho%5Ccos%5Cphi)%5E2&i1=drho%20dphi%20dtheta&i2=0&i3=1&i4=0&i5=pi&i6=0&i7=2pi&podSelect=&showAssumptions=1&showWarnings=1

Answer 19

Yeah similarly it's helpful if you can make the function or shape you're integrating over as symmetric as possible you'll have a better time. Think of it as like a higher dimensional version of how the integral of an even function from -a to +a is the same as 2* the integral from 0 to a or how odd functions will have their own symmetry to exploit.

Answer 20

idk why that link is messed up, but incase you're not familiar with it, just so you know... you may enter the integrand/bounds manually in that link for evaluating the integral in spherical coordinates
http://i.gyazo.com/dd083309ef5104820373658aa5f4ad46.png