Volume of a Circle - Finding r with Double Integrals

[Ry122]

http://users.on.net/~rohanlal/circle2.jpg
this is part of the solution to finding the volume of a circle with double integrals.
I just want to know where the r from rdrd0 came from and also
why the limits on the d0 integral are 2pi and 0.

 

[Hootenanny]

http://users.on.net/~rohanlal/circle2.jpg
this is part of the solution to finding the volume of a circle with double integrals.
I just want to know where the r from rdrd0 came from and also
why the limits on the d0 integral are 2pi and 0.

I assume you mean this is part of a question to find the volume of the cylinder created by extruding a circle along the z-axis.

To answer your first question, the integral has be transformed from Cartesian to polar coordinates. Rather than specifying the position of a point in terms of it's (x,y) coordinates, polar coordinates uses (r,Θ), where r is the distance from the origin to the point and Θ is the angle between the radius and the positive x semi-axis. For more information and answers to your subsequent questions see http://mathworld.wolfram.com/PolarCoordinates.html" .

 

[Ry122]

What makes you think its a cylinder?
This is the full solution:
http://users.on.net/~rohanlal/circ3.jpg

 

[HallsofIvy]

http://users.on.net/~rohanlal/circle2.jpg
this is part of the solution to finding the volume of a circle with double integrals.
I just want to know where the r from rdrd0 came from and also
why the limits on the d0 integral are 2pi and 0.

This is the volume of a sphere, not a circle- circles don't have "volume"!

And you should have learned that the "differential of area in polar coordinates" is r dr d\theta when you learned about integrating in polar coordinates. There are a number of different ways of showing that. I recommend you check your calculus book for the one you were expected to learn.

 

[Ry122]

and why is the limit 2pi to 0?

 

[Hootenanny]

and why is the limit 2pi to 0?

http://mathworld.wolfram.com/SphericalCoordinates.html