# Finding volume using triple integrals.

Use a triple integral to find the volume of the solid: The solid enclosed by the cylinder $$x^2+y^2=9$$ and the planes $$y+z=5$$ and $$z=1$$
This is how I started solving the problem, but the way I was solving it lead me to 0, which is incorrect. $$\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{1}^{5-y}dzdxdy=\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\left(4-y\right)dxdy=\int_{-3}^3\left[4x-xy\right]_{-\sqrt{9-y^2}}^\sqrt{9-y^2}dy= {8\int_{-3}^3{\sqrt{9-y^2}}dy}-2\int_{-3}^3y{\sqrt{9-y^2}}dy$$
If this is wrong, then that would explain why I'm stuck. If this is correct so far, that's good news, but the bad news is that I'm still stuck. If someone could help me out, that would be wonderful, thanks!

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Whoops! Thanks for catching that. I made the change – Jc E Aug 1 '13 at 21:16
I don't have time to write out a full answer, but the last integral (the one that begins $-2\int_{-3}^3 \cdots \,dy$) should have a $y$ outside of the square root... – apnorton Aug 1 '13 at 21:21
Oh, yeah I forgot to put that in. I have that written down on my paper, but I'm still having troubles integrating... That integral that I just corrected, I know I'll do that by parts, but the other one is giving me trouble. – Jc E Aug 1 '13 at 23:37
Don't do that right-hand one by parts--do a substitution of $u = 9-y^2$. In a few minutes, I may be able to work out the other one for you in an answer--I have to put away some dishes from dinner. :) – apnorton Aug 2 '13 at 0:21
Sorry that took so long. But I answered it, finally. :) – apnorton Aug 2 '13 at 2:22

Ok. So you have the triple integral: \begin{align} \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_1^{5-y} \;dz\;dx\;dy &= \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}4-y\;dx\;dy \\ &=\int_{-3}^34x-xy\Bigg|_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\;dy \\ &=\int_{-3}^38\sqrt{9-y^2}-2y\sqrt{9-y^2}\;dy \\ &= 8\int_{-3}^33\sqrt{1-\left(\frac{y}{3}\right)^2}\;dy-2\int_{-3}^3y\sqrt{9-y^2}\;dy \end{align}

Now, I'm going to break this up. For the left-hand integral, we must use trig-substitution. Let $\cos(t) = \frac{y}{3}$. This implies that $dy = -3\sin(t)\;dt$. The limits of integration change as well, to $t=\arccos\left(\frac{-3}{3}\right) = \pi$ to $t = \arccos\left(\frac{3}{3}\right) = 0$.

So, the integral becomes: \begin{align} 24\int_\pi^0\sqrt{1-\cos^2(t)}(-3\sin t)\;dt &= -72\int_\pi^0\sin^2(t)\;dt\\ &=72\int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\;dt\\ &=36\int_0^\pi1-\cos(2t)\;dt\\ &=36\left(t - \frac{\sin(2t)}{2}\right)\Bigg|_0^\pi\\ &=\boxed{36\pi} \end{align}

Now, for the left-hand integral, we apply $u$-substitution. If we set $u = 9-y^2$, then $du = -2y\;dy$. The limits are transformed to $u = 9-(-3)^2 = 0$ to $u = 9-(3)^2 = 0$

So, the integral becomes: \begin{align} -2\int_{-3}^3y\sqrt{9-y^2}\;dy &= \int_0^0\sqrt{u}\;du\\ &= \boxed{0} \end{align}

Well, that wasn't exciting. :)

So, putting it all together, we end up with:

$$V = 36\pi + 0 = \boxed{36\pi}$$

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Thank you so much!! – Jc E Aug 2 '13 at 13:07

You can also use the Cylindrical Coordinates to find the volume. Take a look at the area in which all solid is being projected on $z=0$ plane. It is really a circle with radii $3$.

So we have the following triple integrals as well:

$$\int_{\theta=0}^{2\pi}\int_{r=0}^3\int_{z=1}^{5-r\sin\theta}~rdz~dr~d\theta=36\pi$$

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Awesome graphics, once again! ;-) – amWhy Aug 2 '13 at 12:38

This was an exercice on triple integrals; but at the end we are inclined to check our result against geometric reasoning. Such reasoning tells us that the given solid is half a cylinder of radius $3$ and height $2\cdot(5-1)=8$. Therefore its volume is $36\pi$.

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