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