Please observe the inconsistency in your comments.
1) I have no clue how to do this.
2) I just dont know how to find the intervals for x y z (in spherical).
The second one is much better. The first one is desparate.
Often, it is instructive to consider level curves. We are working with z > 0, so I'll refer to the spherical portion as "hemisphere" and when I say "cone", I'll mean the opening-upward portion.
Hemisphere: x^2 + y^2 + z^2 = 1
Cone 3*z^2 = x^2 + y^2
Consider z = 0
Hemisphere: x^2 + y^2 = 1
Cone 0 = x^2 + y^2
Okay, we have a unit circle from the hemisphere, but the cone has not contributed.
Consider z = 1
Hemisphere: x^2 + y^2 + 1 = 1 ==> x^2 + y^2 = 0
Cone 3 = x^2 + y^2
The hemisphere is done, and the cone has expanded to a circle of radius sqrt(3). Obviously, this is outside the hemisphere, so we won't be using all of this. It is clear, however, that the desired volume is above the cone and below the hemisphere.
We'll need to know where they intersect.
Hemisphere: x^2 + y^2 = 1-z^2
Cone 3z^2 = x^2 + y^2
Then 1 - z^2 = 3z^2 ==> 1 - 4z^2 = 0 ==> (1-2z)(1+2z) = 0 ==> z = 1/2 or z = -1/2. We're not interested in the negative value.
Checking the Level Curves
Hemisphere: x^2 + y^2 = 1-(1/2)^2 ==> x^2 + y^2 = 1 - 1/4 = 3/4
Cone 3(1/2)^2 = x^2 + y^2 = 3(1/4) = 3/4
Excellent, both produce a circle of radius sqrt(3)/2
Hopefully, by this time we have noticed complete radial symmetry. We may simplify our lives by considering Octant I only and multiplying by 4.
In Spherical Coordinates, (x,y) translate to (r,theta), just like polar coordiantes. The only complication is that phi angle. Relating to cartesian coordinates, phi is zero on the positive z-axis. It increases as one gets away from the z-axis and is pi/2 on the x-y plane. phi stops at pi on the negative z-axis.
We should have this:
\(\displaystyle x \in \left[0,\frac{\sqrt{3}}{2}\right]\) -- Remember the intersection producing identical level curves?
\(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right]\) -- This covers Octant I only.
\(\displaystyle \phi \in [0,??]\) -- Oh, come on, you're not going to make me do ALL the work, are you?
You have enough information to figure this one out. Let's see what you get.
