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Using Gauss Divergence Theorem, evaluate the integral $\int_{S}\int F.\hat n dS$

where $F=(4xz,-y^2,4yz)$ . S is surface of solid bounded by sphere $x^2+y^2+z^2=10$ and paraboloid $x^2+y^2=z-2$ and $\hat n$ is outward unit normal

ATTEMPT

Now i used Gauss Divergence theorem and i am having problem setting up triple integral. I tried in cylindrical and spherical coordinates but solving integral becomes a total mess. My textbook has evaluated line integral along bounding curve $x^2+y^2=1$ and $z=3$ which i donot seem to understand why?

Please help?

Thanks

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By the divergence theorem, your integral equals $$ \iiint_E4z+2y\; dV, $$ where $E$ is the region bounded by the sphere and the paraboloid, i.e., in cylindrical coordinates $$ E=\{(r,\theta,z)|0\le \theta \le2\pi, 0\le r \le 1, r^2+2\le z \le \sqrt{10-r^2}\}. $$ It follows that $$ \iiint_E4z+2y\; dV=\int_0^{2\pi}\int_0^1\int_{r^2+2}^{\sqrt{10-r^2}}(4z+2r\sin\theta)rdzdrd\theta = \frac{19\pi}{3}. $$

The line integral proposed in your textbook makes no sense to me. I understand the line integral is computed along the intersection of the sphere and the paraboloid, but what is the integrand? The only reason why there would be a line integral is if the author used the Stoked theorem, but I don't see how one would proceed, as the surface is closed and therefore has no border.

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  • $\begingroup$ How did you solve out this triple integral. I am stuck on this because of z limits $\endgroup$ – Taylor Ted Jan 4 '16 at 3:58
  • $\begingroup$ Just the dz part $\endgroup$ – Taylor Ted Jan 4 '16 at 4:02
  • $\begingroup$ $\int 4rz+2r^2\sin\theta\; dz=4rz^2/2+2zr^2\sin\theta$ $\endgroup$ – Kuifje Jan 4 '16 at 11:48
  • $\begingroup$ why doesnot double integral has square root of 10-r^2 $\endgroup$ – Taylor Ted Jan 4 '16 at 12:20
  • $\begingroup$ After putting z limits it become unsolvable $\endgroup$ – Taylor Ted Jan 5 '16 at 5:37

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