Integral in cylindrical coordinates

[ravenea]

Homework Statement

I need to calculate the integral where the region is given by the inside of x^2 + y^2 + z^2 = 2 and outside of 4x^2 + 4y^2 - z^2 = 3

Homework Equations

The Attempt at a Solution

So far, I think that in cylindrical coordinates (dzdrdtheta):

0 <= theta <= 2pi
sqrt(3)/2 <= r <= 1
-sqrt(2-r^2) <= z <= sqrt(2-r^2)

Are the bounds for the radius and z correct?

 

[LCKurtz]

Homework Statement

I need to calculate the integral where the region is given by the inside of x^2 + y^2 + z^2 = 2 and outside of 4x^2 + 4y^2 - z^2 = 3

Homework Equations

The Attempt at a Solution

So far, I think that in cylindrical coordinates (dzdrdtheta):

0 <= theta <= 2pi
sqrt(3)/2 <= r <= 1
-sqrt(2-r^2) <= z <= sqrt(2-r^2)

Are the bounds for the radius and z correct?

Do you know what this figure looks like? If you look at the surfaces you should see that you don't want to do the dz integral first, on the inside. Do you see why? And don't forget the $r$ in your cylindrical volume element.

 

[ravenea]

In the order drdzdtheta, i get:

0 <= theta <= 2pi
sqrt(z^2 + 3)/2 <= r <= sqrt(2-z^2)
-1 <= z <= 1

I understand your point, the radius varies from z = -1 to z = 1 because of the hyperboloid, but the exercise is asking me to give the integrals for both orders: dzdrdtheta, drdzdtheta

Thanks for the help.

 

[LCKurtz]

If you look at a cross-section in the $z-r$ plane you will see that $z$ is a two-piece function of $r$ on both the top and bottom. So doing $z$ first will require two integrals for $dz$.