Question

Use polar coordinates to find the volume of the solid where T is the region that lies under the plane 3x+4y+z=12, above the xy-plane, and inside the cylinder x^2+y^2=2x.

Answer 1

How do I set up the integral and find the limits of integration?

Answer 2

To get an idea of where this cylinder is located in the x-y-z space, complete the square first:
\[\begin{align*}x^2+y^2&=2x\\
x^2-2x+y^2&=0\\
x^2-2x+1+y^2&=1\\
(x-1)^2+y^2&=1\end{align*}\]
This means the cylinder is centered along the vertical axis, \((1,0,z)\).

Here's a top-down view of the region:

This region tells you how to set up the limits of integration for \(r\) and \(\theta\). Converting to polar, the circle is defined by
\[\begin{align*}(r\cos\theta-1)^2+(r\sin\theta)^2&=1\\
r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta&=1\\
r^2-2r\cos\theta&=0\\
r(r-2\cos\theta)&=0\\
r-2\cos\theta&=0\\
r&=2\cos\theta\end{align*}\]
From this you know that \(0\le r\le2\cos\theta\) and \(0\le\theta\le2\pi\).

Here's a plot of the plane (in the first octant, but you can extend it in every direction):

The range of \(z\) is pretty simple \(0\le z\le12-3x-4y\), or in polar coordinates, \(0\le z\le12-3r\cos\theta-4r\sin\theta\).

So the volume is given by the integral,
\[V=\iiint_TdV=\int_0^{2\pi}\int_0^{2\cos\theta}\int_0^{12-3r\cos\theta-4r\sin\theta}r\,dz\,dr\,d\theta\]