Triple Integral of y^2z^2 over a Paraboloid: Polar Coordinates Method

[stolencookie]

Homework Statement

Evaluate the triple integral y^2z^2dv. Where E is bounded by the paraboloid x=1-y^2-z^2 and the place x=0.

Homework Equations

x=r^2cos(theta) y=r^2sin(theta)

The Attempt at a Solution

I understand how to find these three limits, -1 to 1 , -sqrt(1-y^2) to sqrt(1-y^2) , 0 to 1-y^2-z^2. I then solved the first integral and got y^2z(1-y^2-z^2) I know I need to convert to polar coordinates. I am able to do that 0 to 2pi, 0 to 1.
I am having trouble converting the equation y^2z(1-y^2-z^2) to polar coordinates.

 

[Mark44]

Homework Statement

Evaluate the triple integral y^2z^2dv. Where E is bounded by the paraboloid x=1-y^2-z^2 and the place x=0.

Homework Equations

x=r^2cos(theta) y=r^2sin(theta)

What are these equations? They aren't the equations to convert from cartesian to polar.

The Attempt at a Solution

I understand how to find these three limits, -1 to 1 , -sqrt(1-y^2) to sqrt(1-y^2) , 0 to 1-y^2-z^2. I then solved the first integral and got y^2z(1-y^2-z^2) I know I need to convert to polar coordinates. I am able to do that 0 to 2pi, 0 to 1.
I am having trouble converting the equation y^2z(1-y^2-z^2) to polar coordinates.

 

[stolencookie]

yes they are I am having trouble converting it do I leave the z that left over in?

 

[Ray Vickson]

Homework Statement

Evaluate the triple integral y^2z^2dv. Where E is bounded by the paraboloid x=1-y^2-z^2 and the place x=0.

Homework Equations

x=r^2cos(theta) y=r^2sin(theta)

The Attempt at a Solution

I understand how to find these three limits, -1 to 1 , -sqrt(1-y^2) to sqrt(1-y^2) , 0 to 1-y^2-z^2. I then solved the first integral and got y^2z(1-y^2-z^2) I know I need to convert to polar coordinates. I am able to do that 0 to 2pi, 0 to 1.
I am having trouble converting the equation y^2z(1-y^2-z^2) to polar coordinates.[/B]

It looks like the x-axis in your problem plays the role usually reserved for the z-axis in similar problems. So, you will find it easier to use un-standard polar coordinates $y = r \cos \theta$ and $z = r \sin \theta$, with volume element $dV = r\, dr \,d\theta \, dx$.

 

[stolencookie]

Thank you :).

 

[Mark44]

They aren't the equations to convert from cartesian to polar.

yes they are

No, they aren't. In the usual conversions, $x = r\cos(\theta)$ and $y = r\sin(\theta)$, not with $r^2$ as you show. Use Ray's advice to treat x as you would normally do for z

 

[stolencookie]

No, they aren't. In the usual conversions, $x = r\cos(\theta)$ and $y = r\sin(\theta)$, not with $r^2$ as you show. Use Ray's advice to treat x as you would normally do for z

that was a mistype .