Evaluate the double integral of cos(x^2+y^2) dydx

Question

The double integral is from -2 to 2 as the outer and -sqrt(4-x^2) to sqrt(4-x^2) as the inner, for cos(x^2+y^2) dydx

Answer 1

We use I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b, Note that the area of integration is the circle of radius 2 centered at the origin. Clearly, this is an ideal case to integrate via polar coordinates. Remembering r dr dθ = dy dx, this is rewritten as

I[0, 2π] I[0, 2] r cos r² dr dθ =

I[0, 2π] 1/2 sin r² E[0,2] dθ =

I[0, 2π] 1/2 (sin 4 - sin 0) dθ =

I[0, 2π] 1/2 sin 4 dθ =

1/2 (sin 4) θ E[0,2π] =

1/2 (sin 4) (2π - 0) =

1/2 (sin 4) (2π) =

π sin 4