Question

Use the Divergence Theorem to evaluate the surface integral

Answer 1

Compute the divergence of
`\vec F`.

`\mathrm{div}(\vec F) = (\partial(2x^3+y^3))/(\partial x) + (\partial (y^3+z^3))/(\partial y) + (\partial(3y^2z))/(\partial z) = 6x^2 + 3y^2 + 3y^2 = 6(x^2+y^2)`

By the divergence theorem, the integral of
`\vec F` across
`S` is equivalent to the integral of
`\mathrm{div}(\vec F)` over the interior of
`S`, so that

`\displaystyle \iint_S \vec F\cdot d\vec S = \iiint_{\mathrm{int}(S)} \mathrm{div}(\vec F)\,dV`

The paraboloid meets the
`x,y`-plane in a circle with radius 3, so we have

`\mathrm{int}(S) = \left\{(x,y,z) \mid x^2+y^2\le3 \text{ and } 0 \le z \le 9-x^2-y^2\right\}`

and

`\displaystyle \iiint_{\mathrm{int}(S)} \mathrm{div}(\vec F) \,dV = \int_(-3)^3 \int_(-√(9-x^2))^(√(9-x^2)) \int_0^(9-x^2-y^2) 6(x^2+y^2)\,dz\,dy\,dx`

Convert to cylindrical coordinates, with

`\begin{cases}x = r\cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \\ dV = dx\,dy\,dz = r\,dr\,d\theta\,d\zeta\end{cases}`

so that
`x^2+y^2=r^2`, and the domain of integration is the set

`\left\{(r,\theta,\zeta) \mid 0 \le r \le 3\text{ and } 0 \le \theta\le2\pi \text{ and } 0 \le \zeta \le 9-r^2\right\}`

Now compute the integral.

`\displaystyle \int_0^3 \int_0^(2\pi) \int_0^(9-r^2) 6r^2\cdot r\,d\zeta\,d\theta\,dr = 12\pi \int_0^3 \int_0^(9-r^2) r^3\, d\zeta \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \int_0^3 r^3 (9 - r^2) \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \int_0^3 (9r^3 - r^5) \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \left(\frac94 r^4 - \frac16 r^6\right)\bigg|_0^3 = 12\pi \left(\frac94\cdot3^4-\frac16\cdot3^6\right) = \boxed{729\pi}`