By transforming to polar form, evaluate the following integral:
Int(2,3)Int(0,sqr(4-x^2)) Sin(x^2+y^2 )dydx
2007-08-14 05:46:10 · 3 answers · asked by leigh w 1 in Science & Mathematics ➔ Mathematics
The answer is 0.
Are You sure the domain is this: 2 ≤ x ≤ 3; 0 ≤ y ≤ sqrt(4-x^2)?
Then it contains a single point - (2, 0) - a single common point of the circle with a radius of 2, centered at the origin and a vertical stripe at a distance 2 from the Y-axis, so integral over a domain with a zero area is 0.
2007-08-14 06:08:40 · answer #1 · answered by Duke 7 · 0⤊ 0⤋
The other two answers are correct, of course. I think probably the limits of integration on the first integral should be from a to b, where -2 <= a < b <= 2. For example if the problem were int(0,2)int(0,sqrt(4-x^2)sin(x^2 + y^2)dydx, the region of integration would be the portion of a circle with center at (0,0), radius 2, in the first quadrant. In polar form, this would be int(0,pi/2)int(0,2)(sin r^2)rdrdt, with t for theta.
2007-08-16 08:35:16 · answer #2 · answered by Tony 7 · 0⤊ 0⤋
I'm wondering if you got the limits right. Sqr(4-x^2) is not a real number for x>2, so I don't see how you can integrate the x values from 2 to 3.
2007-08-14 13:07:51 · answer #3 · answered by Math Nerd 3 · 0⤊ 0⤋