# Finding the volume bounded by a cylinder and a plane

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?

• The exact region is undeterminate, as the sign of $z$ is free. – Yves Daoust Dec 11 '18 at 10:06
• The title is incorrect. The volume is bounded by four planes. (Unless I am missing something.) – Yves Daoust Dec 11 '18 at 10:07

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.

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.

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.

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.

• What is that even supposed to mean? – ra1nmaster Nov 14 '15 at 20:46
• 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. – 9301293 Nov 14 '15 at 20:52