# Multiple Integration Volume

1. Aug 1, 2006

### JaysFan31

Compute the volume of the region under the plane z=2x+3y+30 and over the region in the xy plane bounded by the circle (x^2)+(y^2)=2y.

I tried using polar coordinates, but I have no idea what the appropriate range of theta would be. I'm also confused with the 2y as part of the circle equation. Any help with the answer would be appreciated.

2. Aug 1, 2006

### siddharth

Can you post what you've done till now? It will be easier for us to spot the error and help you.

3. Aug 1, 2006

### Office_Shredder

Staff Emeritus
For the 2y:

x2 + (y2-2y + c) = c

Choose your constant c such that the (y2-2y + c) term is a perfect square

4. Aug 1, 2006

### JaysFan31

Ok. Bear with me with all the typing of integrals.

I tried polar coordinates:

z=2rcos(theta)+3rsin(theta)+30

Not sure about the regions of integration but I must be wrong because I keep getting the wrong answer.

I integrated (2rcos(theta)+3rsin(theta)+30)rdrdtheta
The bounds I used were 0 to 2sin(theta) for dr
0 to pi for dtheta

I get something like 103 for the volume, but my assignment program is telling me that's wrong.

Thanks.

5. Aug 2, 2006

### siddharth

Right till here.
You have to find the limits for
$$\int \int (2r \cos \theta + 3r \sin \theta + 30) r dr d\theta$$

(Click on the LaTeX graphic to see the code. Also, read https://www.physicsforums.com/showthread.php?t=8997" tutorial for more.)

I think you've found your limits of integration correctly. Did you check if you made a numerical error in evaluating the integral?

Last edited by a moderator: Apr 22, 2017
6. Aug 2, 2006

### HallsofIvy

Staff Emeritus
$x^2+ y^2= 2y$ is a circle with radius 1 and center at (0, 1).
It might be a good idea to shift the axes to the center of the circle by replacing y in z= 2x+ 3y+ 30 by y-1: i.e. use z= 2x+ 3y- 3+ 30= 2x+ 3y+ 27. And $\theta$ going from 0 to $\pi$ only goes around half the circle- you have to use 0 to $2\pi$.