AakashSudhakar
  • AakashSudhakar
[Calculus 3] If anyone can help me with the following integral, that'd be great! Thanks!
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AakashSudhakar
  • AakashSudhakar
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.
taramgrant0543664
  • taramgrant0543664
@mathmate might be able to help!
ganeshie8
  • ganeshie8
what does divergence thm say ?

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AakashSudhakar
  • AakashSudhakar
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.
Loser66
  • Loser66
.
ganeshie8
  • ganeshie8
are you using spherical coordinates ?
Empty
  • Empty
At the very least, calculate the divergence. :P
AakashSudhakar
  • AakashSudhakar
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.
Empty
  • Empty
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
AakashSudhakar
  • AakashSudhakar
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?
ganeshie8
  • ganeshie8
\[\begin{align}\iint\limits_{S}\vec{F}\cdot d\vec{S} &= \iiint\limits_D \text{div}(\vec{F})\, dV\\~\\ &= \iiint\limits_D 9x^2+9y^2+3z^2-6z^2\, dV\\~\\ &= \iiint\limits_D 9(x^2+y^2+z^2)-6z^2\, dV\\~\\ &= \iiint\limits_D 9\rho^2-6(\rho\cos\phi)^2\, dV\\~\\ \end{align}\]
AakashSudhakar
  • AakashSudhakar
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).
ganeshie8
  • ganeshie8
translating to spherical and simplifying the trig should give you the same expression
AakashSudhakar
  • AakashSudhakar
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!
ganeshie8
  • ganeshie8
setting up the integral, bounds, jacobian is the key. evaluating is not so much important, you may use wolfram to evaluate it
ganeshie8
  • ganeshie8
that 6z^2-6z^2 thingy is just an observation, final answer wont change if that idea doesn't strike to you
AakashSudhakar
  • AakashSudhakar
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.
ganeshie8
  • ganeshie8
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
Empty
  • Empty
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.
ganeshie8
  • ganeshie8
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

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