Triple Integral

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    polar triple
  • #1
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.
 
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  • #2
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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.
stolencookie said:

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.
 
  • #3
stolencookie
yes they are I am having trouble converting it do I leave the z that left over in?
 
  • #4
Ray Vickson
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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##.
 
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  • #5
stolencookie
Thank you :).
 
  • #6
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5,418
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
 
  • #7
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 .
 

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