Volume bounded by sphere and cylinder

Question

How do you calculate the volume of the sphere $$x^2+y^2+z^2=a^2$$ completely inside the cylinder $$ x^2+y^2=ax $$ Same question has been asked here Volume bounded by sphere $x^2+y^2+z^2=a^2$ and cylinder $x^2+y^2=a|x|$ However I am not sure i agree with the answer on couple of points 1) The answer assumes that the figure is symmetric on all axes. However, the figure is only on the positive X-axis, so I believe the integral (if correct) has to be multiplied by 4 and not 8 2) Not sure about the limits in the triple integral especially the limits of R. The answer assumes it varies from 0 to a $ cos\theta $. However, I believe that R is sometimes "a" as well when it is going through the top part of the sphere. Appreciate your help if i have missed something in the answer that is provided (Note unable to comment on the answered question, so posing a separate question)

Answer 1

Assuming wlog $a>0$, for any $z$, $0\le z\le a$ the intersection area between the cylinder and the sphere is given by the intersection area of two circles $S(z)$

• $x^2+y^2=a^2-z^2$
• $(x-\frac a 2)^2+y^2=\frac{a^2}4$

once we have found that area, the volume is given by

$$V=2\int_0^a A(z)dz$$