# Change of Variable Bounds of Integration

One of my practice problems asks us to compute the volume of the region enclosed by the unit sphere $$\{(x,y,z): x^2+y^2+z^2=1\}$$ and the set $$\{(x,y,z): z= |x|\}.$$

My first intuition is to use cylindrical coordinates to preserve my z-coordinate. This gives me $$x^2 + y^2+z^2=1 \implies z = \pm \sqrt{1-r^2}$$ and $$z = |x| \implies z=|r\cos\theta|.$$

From here I have \begin{align*} |r\cos\theta| & = \sqrt{1-r^2}\\ r^2\cos^2\theta & = 1-r^2\\ r & = \pm\frac{1}{\sqrt{\cos^2\theta+1}} \end{align*}

So far I have, as my bounds of integration, $$-\frac{1}{\sqrt{\cos^2\theta+1}}\leq r \leq \frac{1}{\sqrt{\cos^2\theta+1}}$$ and $$0 \leq \theta \leq 2\pi.$$ Where I'm stumped is determining my bounds of integration for $$z$$.

I want to say that it should be $$0 \leq z \leq \sqrt{1-r^2}$$ since we are looking at a region that doesn't fall under negative values of $$z.$$ If I am correct I'm not sure as to why this would be true. Any help would be appreciated.

First, try to understand what your region of integration means. $$z=x$$ is a plane perpendicular to $$y$$ axis. Same is $$z=-x$$. The angle between them is $$90^\circ$$. So draw the projection of your sphere in the $$x-z$$ plane. $$z=x$$ is a line at $$45^\circ$$ to the $$x$$ axis, and $$z=-x$$ is the perpendicular. $$z=|x|$$ is the union of the parts of the lines for $$z\ge0$$. That's either a $$1/4$$ slice of the circle, or the complement. So if you extend to 3D, you will either have $$1/4$$ of the sphere or $$3/4$$ of the sphere. Your problem does not specify which one you need to calculate. So let's assume that you want the smaller part.
Since things happen in the $$x-z$$ plane, use $$y$$ axis as the axis of the cylinder. Then $$r$$ is integrated between $$0$$ and $$1$$, $$\theta$$ between $$\pi/4$$ and $$3\pi/4$$, and the $$y$$ is integrated between $$\pm\sqrt{1-r^2}$$. This is the method of cylindrical shells.