Question

Find the volume of the solid inside of the sphere Formula1" title=" x^2+y^2+z^2 = 4z" alt=" x^2+y^2+z^2 = 4z" align="absmiddle" class="latex-formula"> and above the cone $z = \sqrt{3x^2+3y^2}$.

Answer 1

My approach for this problem is using the washer method. 
Evaluate the circular cross-section of the sphere and cone respectively.
Volume can be represented as sum of circular Areas.
To find volume above the cone it is necessary to subtract volume below cone from total volume of sphere.

$V_{above} = V_{sphere} - \int\limits^b_a {(A_{sphere} - A_{cone})} \, dz$
Area = pi*R^2
We need a function for Radius in terms of z.
If y = 0, then looking at the xz-plane, the radius of the cross-sectional circle for a given z-value will be the x-coordinate.
Solve each equation for x^2. leaving y=0.
Sphere:
$R^2 =x^2 = 4z-z^2$
Cone:
$R^2 = x^2 = \frac{1}{3} z^2$
Last we need limits, these are the points where the cone intersects the sphere.
Set the two Radius functions equal to each other:
$4z - z^2 = \frac{1}{3} z^2$
$z(\frac{4}{3}z-4) = 0$
$z = 0, z = 3$
Now set up the integral
$V = V_{sphere} - \pi \int\limits^3_0 {(4z-z^2)-(\frac{1}{3}z^2)} \, dz$
For volume of sphere we need radius of sphere. By completing the square you can show it is 2.
$x^2 +y^2 +(z-2)^2 = 4 = 2^2$
Now finish integral and get answer for volume above the cone:
$V = \frac{4}{3} \pi 2^3- \pi \int\limits^3_0 {(4z-z^2)-(\frac{1}{3}z^2)} \, dz$
$V = \frac{32}{3} \pi - \pi |^3_0 (2z^2 - \frac{4}{9} z^3)$
$V = \frac{14}{3} \pi$