# Thread: use double integral to find volume of the solid bounded by the paraboloid & cylinder

1. ## use double integral to find volume of the solid bounded by the paraboloid & cylinder

use double integral to find volume of the solid bounded by the paraboloid z=x^2+y^2 above, xy plane below, laterally by circular cylinder x^2 +(y-1)^2 = 1

So, I broke it above and below y-axis, and used polar: r varies from 0 to 2sin(theta) and theta varies from 0 to pi.

V = 2* integral from 0 to pi [[ integral from 0 to 2sin(theta) r^2 * r dr ]] dtheta

first integral results: 8sin^4(theta), second integral results: 3pi.

I think the answer is supposed to be 3pi/2. My limits are right??

checked my calculations:
integrate 2r^2 *r dr from 0 to 2sinx - Wolfram|Alpha
integrate 8sin^4(x) from 0 to pi - Wolfram|Alpha

2. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

The base of the solid is a circle entirely above the x-axis. So what you are doing is correct.

$\int _0^{\pi }\int _0^{2 \text{ sin}(\theta )}r^3drd\theta = \frac{3 \pi }{2}$

3. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

So what you are saying is my limits are correct and the answer should be 3pi/2? So the results I got from wolfram are wrong, since they stipulate the answer is 3pi (I linked my entries)

4. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

Originally Posted by SNAKE
So what you are saying is my limits are correct and the answer should be 3pi/2? So the results I got from wolfram are wrong, since they stipulate the answer is 3pi (I linked my entries)
yes the limits and the answer are correct.
I don't see why you multiply the answer times 2 since there is no above and below (only above the x-axis) unless I am misreading the statement of the problem.
Can you give us the link to where you found the answer 3pi ?

5. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

3pi is what I got when I did it myself (through wolfram):
integrate 2r^2 *r dr from 0 to 2sinx - Wolfram|Alpha
integrate 8sin^4(x) from 0 to pi - Wolfram|Alpha.

The reason I multiply by two is because the limit is from 0 to pi.
The whole thing should be 0 to 2pi for the entire circle that is on the xy plane, no?
I found the answer 3pi/2, but didn't get it myself

6. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

Originally Posted by SNAKE
3pi is what I got when I did it myself (through wolfram):
integrate 2r^2 *r dr from 0 to 2sinx - Wolfram|Alpha
integrate 8sin^4(x) from 0 to pi - Wolfram|Alpha.

The reason I multiply by two is because the limit is from 0 to pi.
The whole thing should be 0 to 2pi for the entire circle that is on the xy plane, no?
I found the answer 3pi/2, but didn't get it myself
The whole thing should be 0 to pi for the entire circle not 0 to 2pi

7. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

but why is it not 0 to 2pi to go around the whole circle? 0 to pi is a semicircle, so that would be half is what ii thought.

8. ## Re: use double integral to find volume of the solid bounded by the paraboloid & cylin

Originally Posted by SNAKE
but why is it not 0 to 2pi to go around the whole circle? 0 to pi is a semicircle, so that would be half is what ii thought.
There is no rule (theorem) in polar coordinates that says 0 to pi is a semicircle and 0 to 2pi describes the whole circle.
In our situation the angle varies from 0 to pi to describe the whole circle once around as follows:
r = 2 sin(angle)
angle 0 gives r = 0
angle 45 degrees gives r = sqrt(2)
90 degrees r = 2
150 degrees r = 1
180 degrees r = 0
and the circle is complete