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

In summary, to evaluate the triple integral y^2z^2dv where E is bounded by the paraboloid x=1-y^2-z^2 and the plane x=0, we can use polar coordinates with the equations y=r*cos(theta) and z=r*sin(theta). By converting to polar coordinates, the limits of integration become 0 to 2pi for theta, 0 to 1 for r, and 0 to 1-y^2-z^2 for the third integral. The equation y^2z(1-y^2-z^2) can also be converted to polar coordinates by substituting y=r*cos(theta) and z=r*sin(theta).
  • #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
stolencookie said:

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
yes they are I am having trouble converting it do I leave the z that left over in?
 
  • #4
stolencookie said:

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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Likes stolencookie
  • #5
Thank you :).
 
  • #6
Mark44 said:
They aren't the equations to convert from cartesian to polar.

stolencookie said:
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
Mark44 said:
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 .
 

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

1. What is a triple integral?

A triple integral is a mathematical concept used in calculus to find the volume of a three-dimensional region. It involves integrating a function over a three-dimensional shape or solid.

2. What are polar coordinates?

Polar coordinates are a system of representing points in a two-dimensional plane using a distance from the origin (r) and an angle from a fixed reference line (θ). They are often used in problems involving circular or symmetric shapes.

3. How do you set up a triple integral in polar coordinates?

To set up a triple integral in polar coordinates, we first need to identify the limits of integration for each variable (r, θ, and φ). The r-limits represent the distance from the origin, the θ-limits represent the angle, and the φ-limits represent the height. We then multiply the function by the appropriate differential elements (r, dθ, and dr) and integrate them from their respective limits.

4. What is a paraboloid?

A paraboloid is a three-dimensional shape that resembles a parabola. It can be formed by rotating a parabola around its axis or by intersecting a cone and a plane.

5. What is the significance of using the polar coordinates method for this triple integral?

The polar coordinates method is particularly useful for solving triple integrals involving symmetric shapes, such as a paraboloid. It simplifies the integration process and can often lead to more elegant solutions compared to using Cartesian coordinates.

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