TITLE: Finding the volume of solid
QUESTION [0 upvotes]: Use triple integral to find the volume of the solid enclosed between the elliptic cylinder $x^2 + 9y^2 = 9$ and the planes $z = 0$ and $z = x + 3$.

REPLY [0 votes]: As with any of these problems, sketch the region.  You should get an elliptic cylinder with the $z$ axis as the axis of symmetry, cut in half diagonally by the given plane.  In fact, you should at once be able to see that the sought-after volume is simply half of the volume of the elliptic cylinder between $z=0$ and $z=6$ - this will serve as a check on your integration result.
As for the integral, which you demand be a "triple" integral (if you get more experienced at this, you'll find that people who actually use these things do not care what you call them), we consider trapezoidal cross-sections parallel to the $y$ axis.  Why?  Because the height of the region is independent of $y$.  At position $x$, the width is $dx$ and the height "varies" between $z=0$ and $z=x+3$.
(I must mention here that this demand for a "triple" integral hobbles the understanding of the problem, in which all we are doing is adding up trapezoidal volume elements, which as you'll see is just a single integral, which is much easier to understand in the first place.)
I hope you see how to derive the limits in $x$ and $y$: simply consider the boundaries of the elliptic cylinder, which do not change with $z$.
The triple integral is thus
$$V = \int_{-3}^3 dx \, \int_{-(1/3) \sqrt{9-x^2}}^{(1/3) \sqrt{9-x^2}} dy \, \int_0^{x+3} dz$$
I hope you see just how silly this is.  The inner two integrals trivially evaluate, and you are left with, as I mentioned above, a single integral:
$$V = \frac{2}{3} \int_{-3}^3 dx \, (x+3) \sqrt{9-x^2}$$
I hope you can evaluate this.  BTW the answer should come out to $9 \pi$.
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