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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.

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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.

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