Usnig clylindrical coordinates to find the volume

[farmd684]

Homework Statement

The solid that is inside the surface r^2+z^2=20 and below the surface z=r^2

Homework Equations

The Attempt at a Solution

I m confused how to solve this problem of get a projection and their intersection. Their intersection gives me a equation of
r^4+r^2-20=0 then i m stuck.

Please help :)

 

[vela]

When you say it's inside the first surface, do you mean it's between z=0 and the surface?

Try factoring r⁴+r²-20 to solve for r².

 

[farmd684]

This is the exact question i have copied from the book.
The solution of the equation gives me r=+-5^1/2 could you please tell me how to proceed with these types of problems.
thanks

 

[vela]

Oh, I just noticed the first surface is a sphere, so talking about its inside isn't vague. Have you sketched the surfaces?

You solved for r incorrectly. You probably got the signs switched when you factored the polynomial.

 

[farmd684]

oh sorry yes i was wrong it should be r=+-2. I m confused about the sketch r^2+z^2=20 is a ellipsoid and z=r^2 is a elliptic paraboloid and i m asked to get the solid that is below the elliptic paraboloid ? If it is how can i get the limits of my integration ?

 

[vela]

The easiest approach would be to calculate the volume of the sphere and then subtract out the volume of the protrusion of the paraboloid into the sphere. You found that the intersection of the two surfaces is a circle of radius 2, so the projection of the protrusion onto the xy-plane is a circle of radius 2 centered on the origin. From this, you can find the limits for r and θ. The limits for z are then given by the two surfaces to find the volume of the protrusion.

 

[farmd684]

Thanks i have got the solution by your method. But i m confused how the book solved the problem. They have formed an integral like this
<br /> \int_0^{2\Pi} \int_0^2 \int_{-{\sqrt{20-r^2}}}^{r^2} rdz dr d\theta + \int_0^{2\Pi} \int_0^{\sqrt{20}} \int_{-{\sqrt{20-r^2}}}^{\sqrt{20-r^2}} rdz dr d\theta

could you please elaborate how this integral is formed.regards

 

[vela]

That's not correct. The lower limit for r in the second term should be 2.

\int_0^{2\pi} \int_0^2 \int_{-{\sqrt{20-r^2}}}^{r^2} r \,dz\,dr\,d\theta + \int_0^{2\pi} \int_2^{\sqrt{20}} \int_{-\sqrt{20-r^2}}^{\sqrt{20-r^2}} r \,dz\,dr\,d\theta

It's not clear to me what's confusing you. It seems if you could solve the problem the other way, you should be able to come up with this solution as well. Do you have a specific question about this expression?