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Integrate the vector function

$$F=(2x)\hat{i}+(4y)\hat{j}-(5z)\hat{k}$$ Over the closed surfaces of the volume defined between the surfaces of

$x^2+y^2+z^2=4$ and $x^2+y^2+z^2=1$ and $z>0$.

I used Green's divergence method but I couldn't able to reach a meaningful solution.

1 Answers 1

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Since you have a closed surface, the divergence theorem would be appropriate.

$\nabla\cdot F = 1$

$\iiint dV = V$

Which is the volume between two hemi-spheres.

$\frac 23 \pi (R^3 - r^3) = \frac {14\pi}{3}$

$\int_0^{2\pi}\int_{0}^{\frac {\pi}{2}}\int_1^2 \rho^2\sin\phi \;d\rho\;d\phi\;d\theta\\\int_0^{2\pi}\int_{0}^{\frac {\pi}{2}}\frac 13 \rho^3\sin\phi| _1^2\;d\phi\;d\theta\\ \int_0^{2\pi}\int_{0}^{\frac {\pi}{2}}\frac 13 (8-1)\sin\phi \;d\phi\;d\theta\\ \int_0^{2\pi}\int_{0}^{\frac {\pi}{2}}\frac 73 \sin\phi \;d\phi\;d\theta\\ \int_0^{2\pi}\frac 73 (-\cos\phi)|_0^{\frac\pi 2} \;d\theta\\ \int_0^{2\pi}\frac 73 \;d\theta\\ \frac 73 (2\pi)\\ \frac {14\pi}3$

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    Hi Doug, Can you explain the mathematical manipulations that you did to solve? Because to do this, in the beginning I differentiated vector function and then it became meaningless and I couldn't do anything. Thanks for help – 2017-01-09
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    $\nabla \cdot F = \frac {\partial}{\partial x}(2x )+ \frac {\partial}{\partial y}(4x)+\frac {\partial}{\partial z}(-5z) = 2+4-5 = 1$ – 2017-01-09
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    Hi again and thanks for your interest, I meant the divergence integration part of your solution. – 2017-01-09
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    I know it is not so hard, also I did the part that you explained but after that I couldn't pass to integration part. I solved x,y,z square examples but I couldn't handle this. – 2017-01-09
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    The point that I couldn't get is how you got 2pi/3 – 2017-01-09
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    You can apply classical geometry from that point (that is what I did). If you want to practice your triple integrations. Convert to spherical $\int_0^{2\pi}\int_{0}^{\frac {\pi}{2}}\int_1^2 \rho^2\sin\phi \;d\rho\;d\phi\;d\theta$ – 2017-01-09
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    I think the result should be 20pi/3 , isn't it? Forgive me for my messages, I'm trying to reach the only right result :) – 2017-01-09
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    How do you get 20? – 2017-01-09
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    But cos(pi/2)=0 and cos(0)= 1 so after second integral it should be 10/3 – 2017-01-09
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    $\frac 73 (-\cos \frac \pi 2 + \cos 0) = \frac 73 (1) =\frac 73$ – 2017-01-09
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    Sorry for my silly fault and appreciate for your patience – 2017-01-09