Use cylindrical coordinates. find the volume of the solid that lies within both the cylinder x2 + y2 = 25 and the sphere x2 + y2 + z2 = 64.

The volume will consist of the cylinder up to where it intersects with the sphere plus the two end caps (top/bottom parts of sphere)

First rewrite equations using cylindrical coordinates:
[tex]r = 5 \\ \\ r^2 +z^2 = 64[/tex]

Find intersection points:
Sub in r =5 into sphere equation
[tex]5^2 + z^2 = 64 \\ \\ z = \pm \sqrt{39}[/tex]

Set up integral:
The general volume for cylinder is [tex]\int \int \int r dr d\theta dz[/tex]

For the inner cylinder you have: 
[tex]V = \int_{-\sqrt{39}}^{\sqrt{39}} \int_0^{2\pi} \int_0^5 r dr d \theta dz \\ \\ =2 \sqrt{39}(2\pi)(\frac{5^2}{2}) \\ \\ = 50 \sqrt{39} \pi[/tex]
It works out the same as cylinder formula [tex]\pi hr^2 = \pi (2 \sqrt{39}) (5^2) = 50 \sqrt{39} \pi[/tex]

For the end caps of sphere, use the triple integral for cylinder volume, however change the limits for 'r' in terms of 'z'.
This is because the radius will be shrinking as 'z' value increases.
we know [tex]r = \sqrt{64-z^2}[/tex] from sphere equation.
Also top/bottom will have equal volumes so you can just double the integral to get volume of both sections.

[tex]V = 2 \int_{\sqrt{39}}^8 \int_0^{2\pi} \int_0^{\sqrt{64-z^2}} r dr d\theta dz \\ \\ =2 \pi \int_{\sqrt{39}}^8 (64-z^2) dz \\ \\ =2 \pi |_{\sqrt{39}}^8 (64z-\frac{z^3}{3}) \\ \\ =2\pi (\frac{1024}{3} - 51\sqrt{39})[/tex]

Combining the volumes yields:
[tex]V = 50\sqrt{39} \pi + \frac{2048}{3} \pi - 102\sqrt{39} \pi \\ \\ V = \frac{2048}{3} \pi - 52\sqrt{39} \pi \\ \\ V = 1124.46[/tex]