Flux outward Sphere-Cylinder

Question

$D=\{x^2+y^2+z^2\le 25,y^2+z^2\le 9\}$

$F=\{y^2,x^2,z\}$

I need to calculate the flux outward the boundary of $D$.

I think I can use the divergent theorem, but How can I define the triple integral of the whole surface? Or Do I need to divide the surface into 3 parts ?

Answer 1

Use the Divergence Theorem. That is, $$ \begin{align*} \iint_E \mathbf{F}\cdot d\mathbf{S} &= \iiint_E \text{div} (\mathbf{F}) \:dV \\ &= \iiint_E \langle \partial_x, \partial_y, \partial_z \rangle \cdot \langle y^2,x^2,z \rangle \: dV \\ &= \iiint_E \partial_x(y^2) + \partial_y(x^2) + \partial_z(z) \: dV \\ &=\iiint_E 1 \:dV \\ &=\iiint_E 1 \:dx \: dy \: dz\\ &=\int_{0}^{2\pi} \int_{0}^{3} \int_{-\sqrt{25-r^2}}^{\sqrt{25-r^2}} \left|\frac{\partial(x,y,z)}{\partial(x,r,\theta)} \right| dx \: dr \: d\theta \\ &=\int_{0}^{2\pi} \int_{0}^{3} \int_{-\sqrt{25-r^2}}^{\sqrt{25-r^2}} \big|r\big| \: dx \: dr \: d\theta \\ &=\int_{0}^{2\pi} \int_{0}^{3} \int_{-\sqrt{25-r^2}}^{\sqrt{25-r^2}} r \: dx \: dr \: d\theta \hspace{4mm} \mbox{ since }r \geq 0 \\ &= 2\pi \int_{0}^{3} r (2\sqrt{25-r^2}) \: dr \\ &= - 2\pi \int_{25}^{16} \sqrt{u}\: du \hspace{4mm} \mbox{ where }u = 25-r^2, \hspace{4mm} \mbox{ so } \hspace{4mm} du = -2r \:dr \\ &= \frac{4\pi}{3}\left( 5^3 - 4^3 \right) \\ &= \boxed{\frac{244}{3}\pi} \\ \end{align*} $$ since given a parametrization in cylindrical coordinates $$ \mathbf{r}(x,r,\theta) = \langle x(x,r,\theta), y(x,r,\theta), z(x,r,\theta)\rangle = \langle x,r \cos \theta,r \sin \theta\rangle, \hspace{4mm} r \geq 0 \hspace{4mm} \mbox{ and } \hspace{4mm} 0\leq \theta \leq 2\pi, $$ its Jacobian is $$ \begin{align*} \frac{\partial(x,y,z)}{\partial(x,r,\theta)} &= \det \begin{pmatrix} \frac{\partial x}{\partial x}& \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial x}& \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial x} & \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} \\ \end{pmatrix} \\ &= \det \begin{pmatrix} 1 & 0& 0 \\ 0 & \cos \theta & -r \sin \theta \\ 0 &\sin \theta & r \cos \theta \\ \end{pmatrix} \\ &= r. \end{align*} $$

Answer 2

You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$\int_{0}^{2\pi}\int_{0}^3\int_{-\sqrt{25-r^2}}^{\sqrt{25-r^2}}r\cdot \text{div}(F)\ dz\ dr\ d\theta$$ Where $r$ is the Jacobian of the transformation $(r,\theta,z)\mapsto (z,r\cos(\theta),r\sin(\theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)