Evaluate $\iint_S\vec{F}$ where $S$ is the surface oriented outwards of the cylinder $x^2+y^2\leq 3$ bounded above and below by two planes.

Question

Use the divergence theorem to evaluate $$\iint_S\vec{F}$$ where $\vec F(x,y,z)=\langle xy^2,x^2y,y \rangle$ and $\vec S$ is the surface oriented outwards of the cylinder $x^2+y^2\leq 3$ bounded above by the plane $z+x=0$ and below by $z=0.$

My solution goes like this:

In order to use, divergence theorem, we must first evaluate $\operatorname{div}(\vec F).$

We note that, $\operatorname{div}(\vec F)=x^2+y^2.$

Now, if $E$ is the solid region corresponding to which $\vec S$ is the boundary then, by divergence theorem we have, $\iint_S\vec{F}=\iiint_{E}\operatorname{div}(\vec F)\, \mathrm{d}V=I.$

We note that, $I=\iint_{D}\int_0^{-x}x^2+y^2\, \mathrm{d}z\, \mathrm{d}A,$ where $D$ is the disk of radius $\sqrt 3$ in the $xy$ plane.

So, $I=\iint_D-x(x^2+y^2)\, \mathrm{d}A=\iint_D-x^3-xy^2\, \mathrm{d}A.$

Converting to polar coordinates, we have, $I=-\int_0^{2\pi}\int_0^{\sqrt 3}r^4\cos^3 t +r^4\sin^2t\cos t\, \mathrm{d}r\, \mathrm{d}t.$

So, $$\begin{align*} I&=-\int_0^{2\pi}\int_0^{\sqrt 3}r^4\cos t(\cos^2 t+\sin ^2 t)\, \mathrm{d}r\, \mathrm{d}t \\ &=-\int_0^{2\pi}\int_0^{\sqrt 3}r^4\cos t\, \mathrm{d}t \\ &=-\frac{9\sqrt 3}{5}\int_0^{2\pi}\cos t\, \mathrm{d}t \\ &=-\frac{9\sqrt 3}{5}[\sin t] \bigg|^{2\pi}_0 \\ &=0 \end{align*}$$

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However, I think my solution is wrong. This is because the answer given is $\frac{18\sqrt 3}{5}.$ I checked my calculations over and over again but was unable to find my mistake. There are two possibilities, either my answer is incorrect or the given answer. Any help regarding this will be greatly appreciated.

Answer 1

If we graph the relevant surfaces,

• $x^2+y^2 = 3$
• $x+z=0$
• $z=0$

the result appears as so:

That is, we have this wedge to contend with:

(Desmos demo.)

The surfaces form $\vec S$, and we'll say the interior is $E$.

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Your calculation of $\operatorname{div} \vec F$ is correct, so we'll forego that.

As for the bounds of integration...

Contrary to the comments, using $z \in [0,-x]$ is correct, provided the other bounds are set up accordingly. If we opt for cylindrical coordinates, we should have $r \in [0,\sqrt 3]$ and $\theta \in [\frac \pi 2, \frac{3\pi}{2}]$, as confirmed by looking at a "top down" view of the wedge:

Correspondingly,

$$\begin{align*} I &= \int_{\theta=\pi/2}^{\theta=3\pi/2} \int_{r=0}^{r=\sqrt 3} \int_0^{-x} (x^2+y^2) \,\mathrm{d}V\\ &= \int_{\theta=\pi/2}^{\theta=3\pi/2} \int_{r=0}^{r=\sqrt 3} \int_0^{-r \cos \theta} r^3 \, \mathrm{d}z\, \mathrm{d}r\, \mathrm{d}\theta \end{align*}$$

Note that the following conversions were used: $$\begin{align*} x &= r \cos \theta \\ x^2 + y^2 &= r^2 \\ \mathrm{d}V &= r \, \mathrm{d}z\, \mathrm{d}r\, \mathrm{d}\theta \end{align*}$$

This integral is fairly standard, so you should be able to work from here. Wolfram confirms the answer to be $$ \boxed{I = \frac{18 \sqrt 3}{5} } $$

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Your bounds of integration are incorrect as they use the entirety of the disk $D$ (note the $[0,2\pi]$ for $\theta$). However, for the other values of $\theta$ I did not use, the planes swap order in terms of being above/below each other (refer to the first graphic) -- it is always important to draw the visuals for these surfaces/solids to get a clean grasp as to what the bounds of integration should be!!