Calculus/Calculating the volume of R^3 lay between the sphere and cilinder
So you have a 3d space where a sphere is present with the equation
A cilinder cuts a certain volume out of this sphere, it's up to me to calculate this volume. The cilinder has cartesian coordinates
I can recognize the equation of a circle (x-x0)²+(y-y0)²=r²
I rewrite it in this form to find the center of the cilinder
so center is (0,a) and radius is a
I go onto polar coordinate system:
So I rewrite the integral
2*integral(0 to pi)integral(0 to 2asin&)sqrt(4a²-p²)pdpd&
-1*integral(0 to 2pi)integral(0 to 2asin&)sqrt(4a²-p²)d(4a²-p²)d&
This gives me
-1*integral(0 to 2pi)[(4a²-p²)^(3/2)/(3/2)]from p=0 to p=2asin&
Up to this I know the calculations are correct. However the following things aren't very clear to me:
Why is that z=sqrt(4a²-x²-y²) only half of the volume is?
Why are the borders of the integral to & readjusted to 2pi during integration?
So now I need to integrate further to &, but there is where I get lost
The result I need to get is (16a^3/6)*(pi-(4/3))
z=sqrt(4a²-x²-y²) is only half of the volume since the cylinder and the sphere
also intersect on the other side (z<0) of the xy-plane.
Your first integral is correct: 2*integral(0 to pi)integral(0 to 2asin&)sqrt(4a²-p²)pdpd&
which gives: (16/9)*a^3*(3*Pi-4)...I think the "6" in your answer should be a "3"...