I'm reviewing for my Calculus 3 midterm, and one of the practice problems I'm going over asks to find the volume of the below solid 1. by using a triple integral with spherical coordinates, and 2. by using a triple integral with cylindrical coordinates.
I'm able to do the integral with spherical coordinates, but I'm getting confused on the one with cylindrical coordinates. In my notes I have written that the cylindrical volume should be: $$dv = r\ dr\ d\theta\ dz$$
Looking at the solution to this problem, it is integrated in the order: $dz\ dr\ d \theta$ , I'm not sure what the point is of changing the integration order?
Also, the bounds in the solution for the integral with respect to $z$ are from $0$ to $\sqrt{4-r^2}$, and I'm not sure where that is coming from either?
Any help would be greatly appreciated.
Best Answer
You know the equation of such part of the sphere is $$z^2=4-(x^2+y^2),x\in[0..2],y\in[0..2]$$ But $r^2=x^2+y^2$ and then $z=\sqrt{4-r^2}$. The ranges of our new variables are : $$\theta|_0^{\pi/2}, r|_0^2, z|_0^{\sqrt{4-r^2}}$$ So we have to evaluate $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}}dv$$