# Homework Help: Triple integral & cylindrical coordinates

1. Dec 15, 2011

### Kuma

1. The problem statement, all variables and given/known data

When you are doing a triple integral and convert it to cylindrical co ordinates, how do you find the new ranges of integration?

I understand the new range of z, if z is between f(x,y) and g(x,y), you just sub in
x = r cos θ and y = r sin θ to find the new functions. But how do I find the range for r and θ? I'm lost there.

Here is a question i was working on and the solution is a bit confusing.

Find the volume of the solid bounded by the paraboloid z = 4x^2 +y^2 and the cylinder
y^2 + z = 2.

So the solution, they replace x = r cos θ / root 2 and y = r sin θ. I'm confused why not just use x = r cos θ? and the range for r is from 0 to 1 and for θ its between 0 and 2pi.

2. Relevant equations

3. The attempt at a solution

2. Dec 15, 2011

### SammyS

Staff Emeritus
Can you show, or explain completely, the triple integral which is given as the solution?

3. Dec 15, 2011

### Kuma

Here is the portion of the solution where they derive the integral.

4. Dec 15, 2011

### vela

Staff Emeritus
Find the equation x and y satisfy where the two surfaces intersect. From that, you might be able to see why x was rescaled the way it was in the solution.

5. Dec 15, 2011

### SammyS

Staff Emeritus
Why does their solution "replace x = r cos θ / root 2 and y = r sin θ" ?

It's as if they first rescale x: Letting x' = (√2)x. Note: x' is not a derivative.

Then x' = r cos(θ) and y = r sin(θ) .

The reason they did this is explained in vela's post.

6. Dec 15, 2011

### Kuma

Thanks.

So its 4x^2 + 2y^2 < 2

But I still don't understand? How would you know to rescale it by multiplying it by 1/sqrt 2?
What does this rescaling do? I'm still a bit lost.

And how do I find the ranges for r and theta?
Does θ vary based on the quadrant or what?

7. Dec 15, 2011

### SammyS

Staff Emeritus
Can you identify what geometric figure the equation 4x2 + 2y2 = 2 describes?

Suppose x' = (√2)x, i.e. x = x'/√2 . Substitute that into this equation. What figure is now described by the result?

8. Dec 15, 2011

### Kuma

It's a circle with r = 1, which I guess is how you get the range of r; with that substitution rather than an ellipse. I wasn't aware you are allowed to just make arbitrary substitutions like that to change the shape.

9. Dec 16, 2011

### SammyS

Staff Emeritus
The Jacobian takes makes up for the distortion that would be caused by the change of coordinates.