Volume of a solid by triple integration

[kasse]

Homework Statement

Find the volume of the solid inside the sphere x^2 + y^2 + z^2 = 4 and over the paraboloid 3z = x^2 + y^2

The Attempt at a Solution

This should be easy to calculate using polar coordinates. The limits for z is [r^2/2, sqrt(4-r^2)] and for tetha: [0, 2*pi], but how do I find the limits for r? The intersection between the two surfaces is: sqrt(4 - r^2) = r^2/3. By inspection I can see that the answer is r = sqrt(3), but what is the mathematical method to find this limit?

 

[Defennder]

You meant cylindrical coordinates, right? Firstly you have the equations r^2 + z^2 = 4 and 3z = r^2. Express z in terms of r and substitute it into the other equation. Then solve for r. You'll get the answer.

 

[HallsofIvy]

Homework Statement

Find the volume of the solid inside the sphere x^2 + y^2 + z^2 = 4 and over the paraboloid 3z = x^2 + y^2

The Attempt at a Solution

This should be easy to calculate using polar coordinates. The limits for z is [r^2/2, sqrt(4-r^2)]

There must be a typo here. You said the parabola was given by 3z= r².

and for tetha: [0, 2*pi], but how do I find the limits for r? The intersection between the two surfaces is: sqrt(4 - r^2) = r^2/3. By inspection I can see that the answer is r = sqrt(3), but what is the mathematical method to find this limit?

Don't look at z, look at z². z= r²/3 implies z²= r⁴/9. So you have r⁴/4/9= 4- r² or r⁴+ 9x- 36= 0.

That can be factored as (r²- 3)(r²+ 12)= 0 so r²= 3 or r^{2= -12. The four roots to that equation are r= \pm\sqrt{3} and r= \pm 2i\sqrt{3}. Of course, r must be positive so r= \sqrt{3}.}