We have the two balls
$x^2 + y^2 + z^2 = 1$
and
$x^2 + y^2 + z^2 - 2z = 0$.
What is the volume of their common part?
I am confused again. My book says $19\pi/12$.
Last night I got the same answer $19\pi/12$ and I was happy.
But upon a second thought, I think it doesn't make sense,
because the volume of a whole ball with radius $1$ is $16\pi/12$.
So there's no way the common part can be $19\pi/12$.
Now I redid the calculations and I am now getting $5\pi/12$.
What is the correct answer here?
How did I solve this?
Both balls have radius 1.
First I found the circle $L$ which is the intersection of the two balls. It is a circle with radius $\sqrt{3}/2$ in the plane $z=1/2$. Right?
Then I projected $L$ onto the $Oxy$ plane. Let's say this projection is the circle $R$, which of course also has radius $\sqrt{3}/2$.
OK, then using the equations of the two balls, the integral boils down to the double integral
$\iint_\limits{R} (z_2(x,y) - z_1(x,y)) \ dx \ dy$
which simplifies to
$\iint_\limits{R} (2\sqrt{1-x^2-y^2} - 1 ) \ dx \ dy$
Then I switched this integral to polar coordinates $(\rho, \phi)$
$0 \le \rho \le \sqrt{3}/2$
$0 \le \phi \le 2\pi$
and I solved it.
And that's how I got $5\pi/12$
Your book is wrong and you are correct.