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Is it fine? Using cylindrical coordinates to compute $$\iiint_T (x^2+y+z^2)^3 dV $$

where $T$ is the solid $x^2+z^2=1$, $y=0$, $y=2$, $z=0$.

enter image description here

Attempt: using cylindrical coordinates $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{1-r^2\cos^2\theta}} (r^2\cos^2\theta + r\sin \theta +z^2)^3\cdot r\cdot dz\ dr\ d\theta $$ I have doubts on the limits of integration, I think $z$ goes from 0 to $\sqrt{1-x^2}=\sqrt{1-r^2\cos^2\theta}$, and $r$ from 0 to 2 and $\theta$ from 0 to $\pi/2$. Is it ok?

apa
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    This problem is better suited for cylindrical coordinates that take the form $$x=r\cos(\theta),y=y,z=r\sin(\theta)$$ If you used cylindrical coordinates of the form $$x=r\cos(\theta),y=r\sin(\theta),z=z$$ you would have to evaluate three separate integrals. – Matthew H. Feb 11 '22 at 16:37
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    What is your solid exactly? Saying you have $x^2+z^2=1$, $y=0$, $y=2$, $z=0$ is quite ambiguous, and doesn't correspond to a solid in $\mathbb{R}^3$. – Lorago Feb 11 '22 at 19:15

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The reason why cylindrical coordiates are natural for this problem is because, as you have noted with your graphed picture of the domain, you are dealing with part of a cylinder. The key thing to notice here, however, is which axis is the axis of rotation. Now I couldn't clearly tell what the region you're saying we are integrating over is, as I think you wrote it in a rather unclear way, but I assume the region we are integrating over is the set

$$\Omega = \{(x,y,z)\in\mathbb{R}^3 : x^2+z^2\leq1 \land 0\leq y\leq 2 \land z \geq 0\}.$$

If this is the case, then you can see from the part $x^2+z^2\leq 1$ that the axis of rotation for the cylinder is the $y$-axis. Therefore it is natural to pick the cylindrical coordinates

$$\begin{cases} x=r\cos \theta, \\ y=y,\\ z=r\sin \theta, \end{cases}$$

as this corresponds to a cylinder with the $y$-axis as its axis of rotation. We now investigate what $\Omega$ corresponds to over this coordinate transform. So we consider the definition of $\Omega$ piece by piece. First we have that

$$x^2+z^2\leq 1 \iff r^2 \cos^2\theta +r^2\sin^2\theta=r^2(\cos^2\theta+\sin^2\theta)=r^2\leq 1,$$

so $0\leq r \leq 1$. Secondly we have that $0\leq y\leq 2$. Lastly we have that (with $r\neq 0$ and as $0\leq \theta\leq 2\pi$)

$$z\geq 0 \iff r\sin \theta\geq 0 \iff 0\leq \theta \leq \pi.$$

Thus our new domain of integration is given by

$$\Gamma=\{(r,\theta,y)\in\mathbb{R}^3 : 0\leq r\leq 1 \land 0 \leq y \leq 2 \land 0\leq \theta \leq \pi\}.$$

We now turn to the actual integral. Using our change to cylindrical coordinates we get that

\begin{align*} I&=\iiint_\Omega (x^2+y+z^2)^3 \mathrm{d}V =\iiint_\Gamma r(r^2\cos^2\theta+y+r^2\sin^2\theta)^3\mathrm{d}V \\ &=\iiint_\Gamma r(r^2(\cos^2\theta+\sin^2\theta)+y)^3\mathrm{d}V =\iiint_\Gamma r(r^2+y)^3\mathrm{d}V. \end{align*}

Finally we can turn this triple integral into an iterated integral, yielding that

\begin{align*} I &= \int_0^1 \int_0^2 \int_0^\pi r(r^2+y)^3 \mathrm{d}\theta~\mathrm{d}y~\mathrm{d}r = \pi\int_0^1 \int_0^2 r(r^2+y)^3 \mathrm{d}y~\mathrm{d}r \\ &= \pi\int_0^1 \biggl[ \frac{r(r^2+y)^4}{4} \biggr]_{y=0}^{y=2}~\mathrm{d}r = \pi\int_0^1 \frac{1}{4}\left(r(r^2+2)^4-r^9 \right)~\mathrm{d}r \\ &= \frac{\pi}{4}\int_0^1 \left(r(r^8+8r^6+24r^4+32r^2+16)-r^9 \right)~\mathrm{d}r \\ &= \frac{\pi}{4}\int_0^1 \left(r^9+8r^7+24r^5+32r^3+16r-r^9 \right)~\mathrm{d}r \\ &= \frac{\pi}{4}\int_0^1 \left(8r^7+24r^5+32r^3+16r \right)~\mathrm{d}r \\ &= \frac{\pi}{4} \biggl[ r^8+4r^6+8r^4+8r^2 \biggr]_0^1 \\ &= \frac{\pi}{4} (1+4+8+8) = \frac{21\pi}{4}. \end{align*}

Thus, if I understood your region correctly, we have that $I=\frac{21\pi}{4}$

Lorago
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