Volume integral between sphere r = 1 and z=x^2

[Moopcan]

Homework Statement

Set up the volume integral of x^2 + y^2 + z^2 between z=sqrt(1-x^2-y^2) and z=x^2

Homework Equations

Cartesian differential volume element, spherical differential volume element

The Attempt at a Solution

In class I've been told when doing triple integrals to try and find an extrusion so that I can reduce the problem to a double integral. I don't see how I can do that in this case though because when looking from the top,. I have an idea of what the problem looks like, it's half a sphere being cut out by a parabolic sheet.

If I try to look on the x-z plane, I can start bounding y(as in the inner integral) between -sqrt(1-x^2-z^2) and sqrt(1-x^2-z^2). But if I try to bound z, I have z < sqrt(1-x^2-y^2), which depends on y and I don't think that really makes sense...

I considered going to spherical at this point because every way I tried to bound it in cartesian led to integration bounds being in terms of previously eliminated variables.

So in spherical, the sphere half becomes r=1 and the parabola sheet becomes r=cot(phi)sec(theta).

theta is between 0 and 2Pi. It's half of a sphere, so phi is between 0 and pi/2. The original function goes to r^2 in spherical and r is between cot(phi)sec(theta) and r=1. The integrand is r^4 * sin(phi) and the order of integration is dr dtheta dphi

I wasn't sure of this so I put it into MathCAD to see if it spat something out that made sense and...it sat there and didn't stop calculating. Which tells me that I've done something wrong.

I get the feeling that something is wrong with my radius limit. And that I'm missing something in cartesian.

I just randomly had another idea. Could I do inner as x^2 < z < sqrt(1-x^2-y^2), intersect the two to yield x^4 = 1 - x^2 - y^2 which when solving for x, yields x^2 = -1/2 + sqrt(5-4*y^2), which gives a really disgusting expression for x in terms of y. Could I use -disgusting expression < x < disgusting expression as my middle and -1<y<1 as the outer?

 

[mighty2000]

Thank you for your post. It seems like you have made some good progress in setting up the volume integral for this problem. I would like to offer some suggestions and clarifications that may help you in finding the correct solution.

Firstly, when dealing with triple integrals, it is important to consider the order of integration. In this case, it may be helpful to start by integrating with respect to z first, then y, and finally x. This way, you can avoid the issue of having bounds for z that are dependent on the other variables.

Next, when setting up the bounds for the integral, it may be helpful to think about the geometry of the problem. As you noted, the region is half of a sphere cut out by a parabolic sheet. This means that the bounds for x and y will depend on the radius r, which is bounded by 1 on the top and by the parabolic sheet on the bottom. So, the bounds for x and y will be -sqrt(1-r^2) < x < sqrt(1-r^2) and -sqrt(1-r^2-x^2) < y < sqrt(1-r^2-x^2).

Finally, when setting up the integral in spherical coordinates, it is important to remember that the function needs to be written in terms of the spherical coordinates, which are r, θ, and φ. In this case, the function is x^2 + y^2 + z^2, which can be rewritten as r^2sin^2φcos^2θ + r^2sin^2φsin^2θ + r^2cos^2φ. This way, you can correctly set up the integrand as well as the bounds for r, θ, and φ.

I hope this helps and good luck with your problem. Remember to always consider the geometry and the order of integration when setting up volume integrals.