Original question: Triple integral in different coordinate systems.
How to change to spherical coordinates in this integration:
$$ \int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{3-\sqrt{13-x^2-y^2}}^{-3+\sqrt{25-x^2-y^2}} \, dz\,dy\,dx, $$
I know this question has been asked in the original post but I didn't quite get how to change here.....
https://i.stack.imgur.com/vykIu.png
So in the figure I am guessing the volume we are calculating is of the hemisphere bounded by the plane of intersection of the sphere and lower portion of the upper sphere.
If that's the case then :
all $y$ values belong to $x^2+y^2\le 9$ all $x$ values same : $Z$ values if I am correct varies from the lower portion of the upper sphere to $z={}$ point of intersection.
Now how do I set up the spherical coordinates using this. What are the steps to analyse them?
I have referred to some examples in my book but I haven't got any clear idea on how to set them up so please help me.
As seen from the cross-section diagram, the volume is of a lens formed from the surfaces of two spheres. Given the vertical orientation or the symmetry around the $z$-axis, it is more convenient to set up the volume integral in cylindrical coordinates, or simply to use the shell integral,
$$V = \int_0^{2\pi}\int_0^3 \int_{3-\sqrt{13-r^2}}^{-3+\sqrt{25-r^2}} dz\> rdr d\theta$$ $$ = 2\pi\int_0^3 \left(\sqrt{25-r^2}+\sqrt{13-r^2}-6 \right)rdr$$
which can be carried out by hand.
Edit: responding to comments
Note that the equations for the two spheres are $r^2+(z-3)^2=13$ and $r^2+(z+3)^2=25$. The limits along the $z$ direction are provided by the surface equations of the two spheres, i.e. $z_1 = 3-\sqrt{13-r^2}$ and $z_2 = -3+\sqrt{25-r^2}$.
Also note the two spheres intersect at $r=3$, which can be obtained by solving the system of equations below, $$r^2+(z-3)^2=13,\>\>\>\>\>r^2+(z+3)^2=25$$ So, use $r=3$ to set the limit for the $r$-integral.