Creating an integral to represent the volume of the intersection of two balls in cartesian coordinates The question states:
Let $A$ be the intersection of the balls
$x^2+y^2+z^2\leq 9$ and $x^2+y^2+(z-8)^2\leq 49$
I am asked to just set up the iterated triple integral that represents the volume of $A$ in cartesian coordinates.
What I am able to determine so far is that for the equation
$x^2+y^2+z^2\leq 9 $:
$-\sqrt{9-x^2-y^2}\leq z \leq \sqrt{9-x^2-y^2}$
$-\sqrt{9-x^2}\leq y \leq \sqrt{9-x^2}$
$-3\leq x \leq 3$
If I were to set up this integral it would be:
$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}} dzdydx$
But I don't know how I'm supposed to set up the intersection of the two balls?
I was thinking of setting the two equations equal to each other so that
$x^2+y^2\leq 9-z^2$ and
$x^2+y^2\leq 49-(z-8)^2$
so then I have $9-z^2=49-(z-8)^2$
Solving for $z$ I get $z=3/2$ but I don't know what to do with this information.

$\def\bbr{\mathbf{R}}$It is simpler to exploit the symmetries. From the picture below, you can see clearly that the intersection is a normal domain in $\bbr^3$, which means that you can write it as
$$
A=\{(x,y,z)\mid (x,y)\in D, f(x,y)\le z\le g(x,y)\}
$$
where $D$ is some two-dimensional region on the $xy$-plane. You can then set up the volume as
$$
V=\iint_D\int_{f(x,y)}^{g(x,y)}1\;dz dA
$$
and reduce it further to an iterated integral by working on the region $D$.

The ball $x^2+y^2+z^2\le 9$ can be written as:
$$
E_1=\{(x,y,z)\mid (x,y)\in D_1,-\sqrt{9-x^2-y^2}\le z\le\sqrt{9-x^2-y^2}\}
$$
where $D_1=\{(x,y)\mid x^2+y^2\le 9\}$ is a 2D region.
Similarly, the ball $x^2+y^2+(z-8)^2\le 49$ can be written as
$$
E_2=\{(x,y,z)\mid (x,y)\in D_2,8-\sqrt{49-x^2-y^2}\le z\le 8+\sqrt{49-x^2-y^2}\}
$$
where $D_2=\{(x,y)\mid x^2+y^2\le 49\}$. So
$$
E_1\cap E_2=\{(x,y,z)\mid (x,y)\in D, 8-\sqrt{49-x^2-y^2}\le z\le\sqrt{9-x^2-y^2} \}\tag{1}
$$
for $D=\{x^2+y^2\le R^2\}$ with some real number $R$. To find out $R$, observe that the intersection curve (see picture below) of the two spheres is given by the equations:
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
x^2+y^2+(z-8)^2= 49
\end{cases}
$$
which is equivalent to
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
9-z^2+(z-8)^2= 49
\end{cases}
$$
which is
$$
\begin{cases}
x^2+y^2+z^2= 9 \\
z=\frac{3}{2}
\end{cases}
$$
So $R = \frac{3\sqrt{3}}{2}$. Now you can write the volume as
$$
V=\iint_D\int_{8-\sqrt{49-x^2-y^2}}^{\sqrt{9-x^2-y^2}}1\cdot dz dA
=\int_{-R}^{R}\int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}}\int_{8-\sqrt{49-x^2-y^2}}^{\sqrt{9-x^2-y^2}}1\cdot dzdxdy
$$

