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I have been given the following equations:

$$x^2 + z^2 = 9$$ $$ x = 0 $$ $$ y = 0 $$ $$ z = 0 $$ $$ x + 2y = 2 $$

and have been asked to find the volume of the bounded region. I understand the principle of the integration and how to apply it in this scenario, but I tend to have trouble determining the bounds of integration for a three dimensional region such as the one described above. I have tried sketching the cylinder and the plane, but to no avail; I'm consistently unable to determine the bounds. In general, how would one go about determining the integral bounds for a region such as the above?

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  • $\begingroup$ The exact region is undeterminate, as the sign of $z$ is free. $\endgroup$ – Yves Daoust Dec 11 '18 at 10:06
  • $\begingroup$ The title is incorrect. The volume is bounded by four planes. (Unless I am missing something.) $\endgroup$ – Yves Daoust Dec 11 '18 at 10:07
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Sketching is indeed a good idea when you deal with non-trivial boundaries. In this case, the main boundaries are the coordinate planes $x=0$, $y=0$, and $z=0$. Then, by inspection, we notice that the given plane $x = 2- 2y$ makes the bound region (in $x$-$y$ plane) in the first octant, which means that our boundaries become

$x\in[0, 2-2y]$, $y\in [0, 1]$ and $z \in [0, \sqrt{9-x^2}]$.

The latter boundary appears due to the fact that the cylinder is oriented along the $z$-axis.

I guess practice of working with this type of problems bring essential experience and then each new problem will look easier and easier.

Hope this helps.

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OP: Just to provide some perspective on what others (such as @Pavel) have said:

The desired volume is bounded by the planes $x=0$, $y=0$ and $z=0$. Since the equation of the cylinder is given by $x^2+z^2=9$, this means that the axis of the cylinder is parallel to the $y$ axis, and it has a radius $r=3$.

Now, $x+2y=2$ is the equation of a plane that lies parallel to the $z$ axis, and hence does not depend on $z$. Also, where this plane intersects the $xy$ plane, the equation of the line (in the $xy$ plane) is $y=\frac{1}{2}\left(2-x\right)$. Also, since this plane lies in the positive quadrant, it can be inferred that $x \ge 0$ and $y \ge 0$ is desired. Finally, because of the symmetry of the problem, I can assume $z \ge 0$.

Therefore, the desired region is bounded above by the cylinder $z=\sqrt(9-x^2)$ and below by the plane $z=0$. Also, projecting into the $xy$ plane, the area is bounded by $0 \le y \le \frac{1}{2}\left(2-x\right)$, and $0 \le x \le 2$.

$$\therefore V= \int_0^2{\int_0^{\frac{1}{2}\left(2-x\right)}{\int_0^{\sqrt(9-x^2)} {1}\mathrm{d}z}\mathrm{d}y}\mathrm{d}x$$ where $V$ is the desired volume.

enter image description here

As others have pointed out, it is best to be able to sketch these regions (by hand) and use a CAD tool (e.g. Matlab, Mathematica, if available) to confirm correctness. This will build your confidence that subsequent sketches are correct.

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Apologies in advance, for bad English. You need to start off, by integrating left, integrating right and combining. For example, z component vector, you can integrate first in this case (because it is involved in the least equations). Then, you may go about, triple integrate the other ones (for example, x component first then y component). You will easily be able to determine the integral bounds after this, as you've integrated all three components.

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  • $\begingroup$ What is that even supposed to mean? $\endgroup$ – ra1nmaster Nov 14 '15 at 20:46
  • $\begingroup$ Posts on SE use mathjax(link: en.wikipedia.org/wiki/MathJax), please review this and format your answers to use this to answer questions in the future. $\endgroup$ – 9301293 Nov 14 '15 at 20:52

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