Finding the volume - Polar coordinates

[Dilemma]

Hello everyone,

1. Homework Statement
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

Homework Equations

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:

Thanks in advance,

 

[Mark44]

Hello everyone,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

Attempt : 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

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.

. Parabolaid surface's eq. is (1-r²).

What you show here is not an equation. What equation do you get?

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:

Thanks in advance,

 

[Dilemma]

Hello Mark,

I am sorry for the inconvenience I caused.

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?

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

Then, I converted parabolaid's surface equation.

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

 

[Mark44]

Hello Mark,

I am sorry for the inconvenience I caused.

No problem. That's what we're here for.

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

Then, I converted parabolaid's surface equation.

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

You should get a different result from these equations.

 

[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.

 

[Eclair_de_XII]

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

 

[LCKurtz]

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.

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.

 

[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.