Consider a cone centered about the positive z axis with its vertex at origin,a $90^{\circ}$ angle at its vertex,topped by a sphere of radius $6$.Compute the volume of region bounded by sphere and cone.
My problem:I need help about the limits of integration in spherical coordinates.
Thanks.
On
A picture always helps. The green region is the one you are interested in (assuming sphere centered at origin)
The limits of integration for spherical coordinates as you rightly noted in comments in the other post are $0\leq r \leq 6$ and $0\leq \theta \leq 2\pi$
For finding the limits for $\phi$, refer to the picture and note that the problem specified that the angle of the black triangle at the vertex is $90^\circ = \pi/2$. Consider how much of an angle from the red line you will need to go down (hint: the red line bisects the angle) to find the largest $\phi$ can be.
Here's a way to determine the limits. Given the cone's angle, its equation is $x^2+y^2=z^2$. assuming the sphere is centered at the origin, its equation is $x^2+y^2+z^2=R^2$, with $R=6$
In cone-sphere problems, remember that the cone extends forever, but the sphere is bounded. On the other hand, the cone starts as a point. If you take z-slices of the volume, it starts off as a single point (the apex of the cone), but widens thereon.
The slices are essentially circles $x^2+y^2=z^2$ for a fixed 'z'. However, beyond $z=3\sqrt{2}$, the slice at which the cone intersects the sphere, the outer limit is determined by the sphere. They are still bounded by circles, but of radius $\sqrt{R^2-z^2}$.
This should be enough information for you to find the limits.
Edited to add: If you work through, your integral will split into two integrals.
$$ \int_{z=0}^{3\sqrt{2}} \int_{r=0}^{z} \int_{\theta=0}^{2\pi} r d\theta dr dz + \int_{z=3\sqrt{2}}^{6} \int_{r=0}^{\sqrt{6^2-z^2}} \int_{\theta=0}^{2\pi} r d\theta dr dz $$
ETA2: In spherical,
$$ \int_{\theta=0}^{2\pi} \int_{\phi=0}^{\pi/4} \int_{r=0}^{6} r^2 \sin\phi dr d\phi d\theta $$