What is the polar double integral for a given solid?

[jwxie]

Homework Statement

Compute the indicated solid in POLAR COORDINATE using double integrals.

Below z = 4 - x^2 - y^2, z = x^2 + y^2, between y = x and y = 0.

Homework Equations

The Attempt at a Solution

First of all, the integrand is z = 4 - x^2-y^2 which in polar is 4 - r^2

The limit for the region D in polar is the intersections of y = x, y = 0 of the circle. To find that particular circle I think we have to solve the two z equations, which give us x^2 + y^2 = 2 in the end. This is a circle with radius 2

The limit of region D is 0 <= r <= sqrt(2), and for theta (i use x) is 0 <= x < pi/6
I am not sure whether pi/6 is really the intersecting point of y = x on the circle... Please cofirm that...

This will give us the double integrals
integral (0 to pi/6) integral (0 to sqrt(2) (4 - r^2)r dr d theta

I think this give us pi/2 which is right from the book. But the book only gave pi/2 there is no work shown so I can't tell whether my work is right or not.

Please tell me if I am wrong in the limit of integrations.
Thankyou.

 

[Raskolnikov]

Below z = 4 - x^2 - y^2, z = x^2 + y^2, between y = x and y = 0.

Are you sure it's not below z = 4 - r^2...and above z = r^2? Because It doesn't make much sense if it's below both.

First of all, the integrand is z = 4 - x^2-y^2 which in polar is 4 - r^2

mhmm...good so far.

The limit for the region D in polar is the intersections of y = x, y = 0 of the circle. To find that particular circle I think we have to solve the two z equations, which give us x^2 + y^2 = 2 in the end. This is a circle with radius 2

No, sorry this doesn't work. First of all, x^2 + y^2 = 2 is a circle with radius sqrt(2), not 2. Regardless of that, however, you are not finding your region D correctly. The region D is the bounded by the trace of the paraboloid z=4-r^2 onto the xy-plane; i.e., where z=0. Can you figure out what type of curve the paraboloid traces here?

The limit of region D is 0 <= r <= sqrt(2), and for theta (i use x) is 0 <= x < pi/6
I am not sure whether pi/6 is really the intersecting point of y = x on the circle... Please cofirm that...

This will give us the double integrals
integral (0 to pi/6) integral (0 to sqrt(2) (4 - r^2)r dr d theta

I think this give us pi/2 which is right from the book. But the book only gave pi/2 there is no work shown so I can't tell whether my work is right or not.

Please tell me if I am wrong in the limit of integrations.
Thankyou.

The region D is pretty. The limits of r are from 0 to some integer (which I'll leave you to figure out from my notes above). The problem asks for the volume bounded by the two paraboloids between the planes y=x and y=0. You use these two equations to find the limits for theta. Once again, you look at the xy-plane where your region D lies. What is theta when y=0? What is theta when y=x? These two values become your limits for theta.

And, yes, the final answer is pi/2.

I hope this helps. Good luck.

 

[benorin]

Your integrand lacks the lower boundary, and the angle between y=0 and y=x is Pi/4. Fixing these gives Pi/2 as well, the correct answer.