Find volume of the solid bounded by $x^2+y^2+z^2=a^2$; $x^2+y^2>a|x|$

My attempt was to turn the problem into a problem of solving a triple integral. From $x^2+y^2+z^2=a^2$; $x^2+y^2>a|x|$ I get that $a^2>a|x|+z^2$ and $z^2<a^2-a|x|$. Then from $x^2+y^2>a|x|$ I get $y^2>a|x|-x^2$ and from here $a|x|-x^2>0$ and so $-a < x < a$. But I can't find an upper bound for $y$ relative to $x$. My intention is to create a triple integral and integrate $f(x,y,z)=1$. Any suggestions? Also, I have seen that such problems are solved by solving a double integral instead. What is the approach to transform the problem into a problem of solving a double integral? I'm having trouble with this kind of problems and it isn't clear what approach I should follow. Thanks in advance.

Solution 1:

The region is inside the sphere $x^2 + y^2 + z^2 = a^2$ and for $a \gt 0$, outside the cylinders $x^2 + y^2 = ax, x \geq 0$ and $x^2 + y^2 = - ax, x \leq 0$

It is easier to switch to cylindrical coordinates. Also you state that $f(x, y, z) = 1$. As you are finding volume, note the volume for both $x \geq 0$ and $x \leq 0$ is same. So you can choose to just evaluate one integral and multiply the result by $2$.

In cylindrical coordinates, $x = r \cos\theta, y = r\sin \theta, z = z$

So $~x^2 + y^2 \geq ax \implies r \geq a \cos\theta$. As $x \geq 0, - \pi/2 \leq \theta \leq \pi/2$. Also the projection of the sphere in xy-plane is a circle of radius $a$ which gives the upper bound of $r$. The bounds of $z$ is simply the lower and upper half of the sphere. So the integral to find volume is,

$V = \displaystyle 2 \int_{-\pi/2}^{\pi/2} \int_{a\cos\theta}^a \int_{- \sqrt{a^2 - r^2}}^{ \sqrt{a^2-r^2}} r ~ dz ~ dr ~ d\theta $

Or, $V = ~\displaystyle 4 \int_{-\pi/2}^{\pi/2} \int_{a\cos\theta}^a r ~ \sqrt{a^2 - r^2} ~ dr ~ d\theta $