- #1
DieCommie
- 157
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I am having trouble with this seemingly easy problem.
Evaluate the double integral (sin(x^2+y^2)) , where the region is 16=<x^2+y^2=<81.
I find the region in polar coordinates to be 4=<r=<9 0=<theta=<2pi
I find the expression to be sin(rcos^2theta+rsin^2theta) r dr dtheta , which is equal to sin(r) r dr dtheta
Is what I have done right? I evaluate the integral, first with respect to r over using 4 and 9, then with respect to theta using 0 and 2pi. The answer I get is ~42.43989 but that is wrong
any ideas, hints would be greatly appreciated thanks
Evaluate the double integral (sin(x^2+y^2)) , where the region is 16=<x^2+y^2=<81.
I find the region in polar coordinates to be 4=<r=<9 0=<theta=<2pi
I find the expression to be sin(rcos^2theta+rsin^2theta) r dr dtheta , which is equal to sin(r) r dr dtheta
Is what I have done right? I evaluate the integral, first with respect to r over using 4 and 9, then with respect to theta using 0 and 2pi. The answer I get is ~42.43989 but that is wrong
any ideas, hints would be greatly appreciated thanks