Volume within the sphere Find the volume of the solid that lies within the sphere ,
 above the xy plane, and outside the cone 
My problem is finding the integral function and the limits 
 A: First, here is a sketch showing that the cone cuts out washers at each value of $z$.  The inner radius is given by the cone and the outer radius by the sphere.  In cylindrical coordinates, the cone is given by $z = 7r$ and the sphere by $r^2 + z^2 = 1 \rightarrow r^2 = 1 - z^2$.  The $z$ coordinate goes from $z = 0$ to whatever value of $z$ makes the radius of the cone equal to the radius of the sphere (at that $z$ value):
$$
\text{cone}: r = \frac{z}{7} \text{ plug into sphere equation}\\
\left(\frac{z}{7}\right)^2 + z^2 = 1 \rightarrow z^2 = \frac{1}{1 + \frac{1}{49}} = \frac{49}{50} \\
z_{top} = \frac{7}{\sqrt{50}}
$$
Now you just need to find the volume of each differential washer:
$$
dV = Adh = \pi(R^2 - r^2)dz = \pi\left((1 - z^2) - \left(\frac{z}{7}\right)^2\right)dz
$$
The outer radius, $R$, is from the sphere and the inner radius, $r$, is from the cone.  So finally, you get the volume:
$$
V = \int dV = \pi\int\limits_0^{\frac{7}{\sqrt{50}}}\left(1 - \frac{50}{49}z^2\right)dz = \pi\left.\left(z - \frac{50}{147}z^3\right)\right|_0^{\frac{7}{\sqrt{50}}} \\
V = \pi \left(\frac{7}{\sqrt{50}} - \frac{50}{3\cdot49}\cdot\frac{7^3}{50\sqrt{50}}\right) \\
V = \pi\left(\frac{7}{\sqrt{50}} - \frac{7}{3\sqrt{50}}\right) = \frac{7\pi}{\sqrt{50}}\cdot \frac{2}{3} = \frac{14\pi}{3\sqrt{50}}
$$
edit
If you absolutely must, the "full" integral, in cylindrical coordinates, would be something like this:
$$
V = \int\limits_0^\frac{7}{\sqrt{50}}dz\int\limits_{\frac{z}{7}}^{\sqrt{1 - z^2}}dr\int\limits_0^{2\pi} rd\phi
$$
...but it's easier to visualize it in this case and just use high school geometry to create a single integral.
