Finding the volume - Polar coordinates

1. Jan 2, 2017

Dilemma

Hello everyone,

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

Question : Find the volume of the region which remains inside the cyclinder x 2 + y 2 = 2y, and is bounded from above by the paraboloid surface x 2 + y 2 + z = 1 and from below by the plane z = 0

2. Relevant equations

3. The attempt at a solution

This looks like a pretty straightforward question. First thing to do is converting equations to polar coordinates. When they are converted, the cylinder's region can be written as r = 2sinθ. Parabolaid surface's eq. is (1-r2).Through these, a double integral can easily be set up as the following:

∫(from 0 to π)∫from 0 to 2sinθ) (1-r2)rdrdθ.

However, the solution to this problem proposes a different approach which does not return the same answer.

Here it is:

Last edited: Jan 2, 2017
2. Jan 2, 2017

Staff: Mentor

No.
If you convert the cylinder's equation to polar coordinates, you get $r^2 = 2r\sin(\theta)$, which is different from what you show.
What you show here is not an equation. What equation do you get?

3. Jan 2, 2017

Dilemma

Hello Mark,

I am sorry for the inconvenience I caused.

I simplified $r^2 = 2r\sin(\theta)$.

Then, I converted parabolaid's surface equation.

$z = 1 - x^{2}-y^{2}$
$z = 1 - r^{2}$​

4. Jan 2, 2017

Staff: Mentor

No problem. That's what we're here for.
You should get a different result from these equations.

5. Jan 2, 2017

Dilemma

Here is my reasoning :

By polar coordinates definition $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Therefore, $x^{2}+y^{2} = r^{2}$ I do not see any problem. Also, the answer confirms my logic. The only problem is that the answer divides the integral into two, one of them is from 0 to $\pi/6$, and the other is from $\pi/6$ to $\pi/2$. However, second part of the integral's "r" range is 0 to 1, and I am having difficult times understanding this. My coursebook has a similar question solved in a similar manner as mine.

6. Jan 2, 2017

Eclair_de_XII

You know, I feel like you're missing an extra integral sign. Shouldn't you use cylindrical coordinates, or something?

7. Jan 2, 2017

LCKurtz

Look at the first quadrant in this picture, which shows the circle traces in the $xy$ plane:

The brown region is the polar area over which you need to integrate. The $30^\circ$ line divides the area into two parts. In the lower sliver under the line $r$ goes from $0$ to the pink circle. Above the line $r$ goes from $0$ to the brown circle.

8. Jan 2, 2017

Dilemma

Thank you. I now understand that I did not consider the region which is limited by the surface x 2 + y 2 + z = 1.