Question

Verify that the Divergence Theorem is true for the vector field F on the region E. Give the flux. F(x, y, z) = x2i + xyj + zk, E is the solid bounded by the paraboloid z = 25 − x2 − y2 and the xy-plane.

Answer 1

Parameterize the paraboloidal part by

$\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(25-u^2)\,\vec k$

and the planar part by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

both with $0\le u\le5$ and $0\le v\le2\pi$.

For the paraboloidal part, take the normal vector to be

$\vec r_u\times\vec r_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

and for the planar part,

$\vec s_v\times\vec s_u=-u\,\vec k$

The flux over the paraboloidal part (call it $S_1$) is

$\displaystyle\iint_{S_1}\vec F\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^5(u^2\cos^2v\,\vec\imath+u^2\cos v\sin v\,\vec\jmath+(25-u^2)\,\vec k)\cdot(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^5(25u-u^3+2u^4\cos v)\,\mathrm du\,\mathrm dv=\frac{625\pi}2$

and over the planar part (call it $S_2$),

$\displaystyle\iint_{S_2}\vec F\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^5(u^2\cos^2v\,\vec\imath+u^2\cos v\sin v\,\vec\jmath)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

but this integral vanishes because the dot product is 0. So the total flux over $\partial E$ (the boundary of the given region $E$) is $\dfrac{625\pi}2$

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Confirming with the divergence theorem: the divergence of the given vector field is

$\nabla\cdot\vec F=2x+x+1=3x+1$

By the divergence theorem, the flux is equivalent to the volume integral

$\displaystyle\iiint_E(3x+1)\,\mathrm dV$

$=\displaystyle\int_0^{2\pi}\int_0^5\int_0^{25-r^2}(3r\cos\theta+1)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{625\pi}2$

(where the integral was set up with cylindrical coordinates)