# use a double integral to find volume of the solid cut out of the sphere x^2+y^2+z^2=16 by the cylinder x^2+y^2-4x=0?

Update:

I came up with 512/9, can anyone confirm this answer? Please show work.

Relevance
• Volume = twice the double integral of sqrt(64-r^2) r drdt, r from 0 to 3 and t from 0 to 2pi. V = 4pi/3 * (64^(3/2) - 55^(3/2)) = 4pi/3 * (512 - 55sqrt(55)) ~ 436.09

• I was astonished to find 2 questions about the same solid yesterday and today - the solid, bounded by

the sphere x² + y² + z² = R² and the cylinder x² + y² = Rx (R = const > 0),

is called Viviani's solid (Viviani's window). It is shown here:

http://en.wikipedia.org/wiki/Viviani%27s_curve

http://mathworld.wolfram.com/VivianisCurve.html

As to the volume V, it can be expressed by double integral

V = 2 ∫ ∫ [D] √(R² - x² - y²) dx dy

where D is the circle (projection onto xy-plane): x² + y² = Rx, or

D: π/2 ≤ θ ≤ π/2, 0 ≤ ρ ≤ R cosθ

The integral gives the volume of the upper half of the solid (above xy-plane), so it is taken twice above. In polar coordinates it is expressed by the repeated integral

V = 2 ∫ [-π/2..π/2] dθ ∫ [0..R cosθ] √(R² - ρ²) ρ dρ =

= 4 ∫ [0..π/2] dθ ∫ [0..R cosθ] √(R² - ρ²) ρ dρ =

= (4/3)R³ ∫ ∫ [0..π/2] (1 - sin³θ) dθ = (2π/3 - 8/9) R³

As I can understand from your work, you have in view the remaining solid after removing from the right hemisphere (whose volume is 2πR³/3) the Viviani's solid - its volume is exactly (8/9)R³ = 512/9 if R = 4.