Find volume using double integrals? Question: Use double integral to find the volume of the solid enclosed by the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+(z-1)^2=1$
Alright so I tried to doing this by myself and I'm not sure if this is right. Could someone check over my work?
Curve of intersection:
\begin{align*}
x^2 + y^2 + z^2 &= x^2 + y^2 + (z - 1)^2\\
\implies z^2 &= z^2 - 2z + 1\\
\implies z &= 1/2.
\end{align*}
So, the curve of intersection is $x^2 + y^2 = 3/4$ with $z = 1/2$.
By symmetry, it suffices to double the volume between $z = 1/2$ and $z = 1$.
$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.
Solving for (positive) $z$ yields
$z = \sqrt{1 - x^2 - y^2}$ and $z = 1 + \sqrt{1 - x^2 - y^2}$.
Hence, the volume equals
\begin{align*}
V &= 2  \iint \left[(1 + \sqrt{1 - x^2 - y^2}) - \sqrt{1 - x^2 - y^2}\right]\, dA\\ \\
&= 2 \int_0^{2\pi}\!\int_0^{\sqrt{3/4}}\!r\,dr \,d\theta, \qquad\textrm{via polar coordinates}\\ \\
&= 2 \int_0^{2\pi} \!\left[(1/2)r^2\right]_{r = 0}^{\sqrt{3/4}}\, d\theta\\ \\
&= \int_0^{2\pi}\!3/4\,d\theta\\ \\
&= 3\pi/2.
\end{align*}
 A: By the word "enclosed", I am assuming you want to find the volume bounded by these two spheres, which means the solid formed by their intersection. You claimed:

$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.

If we are truely like to find the volume of the intersection, then this is not accurate. First sphere is centered at $(0,0,1)$ with radius $1$, second has origin as its center with the same radius. If you imagine a viewer sitting on the $x$-axis ($y=0$), this viewer will see the  $x^2 + y^2 + (z-1)^2 = 1$'s bottom lies beneath the top of $x^2 + y^2 + z^2 = 1$, the integral should be set as:
$$
V = \int^{\sqrt{3}/2}_{-\sqrt{3}/2} \int^{\sqrt{3/4 - y^2}}_{-\sqrt{3/4 - y^2}} 
\int^{\sqrt{1 - x^2 - y^2}}_{1- \sqrt{1 - x^2 - y^2}} 1\,dz\,dx\,dy.
$$
Changing to cylindrical coordinates: 
$$
V = \int^{2\pi}_{0} \int^{\sqrt{3}/2}_{0} 
\int^{\sqrt{1 - \rho^2}}_{1- \sqrt{1 - \rho^2}} \rho\,dz\,d\rho\,d\theta.
$$
Integrate above (for you to try):
$$
V = \frac{5\pi}{12},
$$
which coincides with the spherical cap formula. (EDIT: computation error earlier, now fixed)
A: Besides to @Shuhao's approach you can use the Spherical coordinates as well. In fact the limits in this system would be:
$$r|_{\frac{1}{2\cos(\theta)}}^1,~~\theta|_0^{2\pi},~~\phi|_0^{\pi/3}$$
Note that two creatures intersects at $z=1/2$. 

