1. Compute the double integral ∫∫R sin(x^2) dA where R is the triangle with vertices (0,0), (0, sqrt(pi) ), ( sqrt(pi), sqrt(pi) )

2. Compute the double integral ∫∫R sin^6 (x^2 + y^2) dA over the ring between the circles x^2 + y^2 = pi and x^2 + y^2 = 2pi

For both problems, let's make a change of variables : use polar coordinates ( r , t ).

x = r.cos(t)
y = r.sin(t)
x² + y² = r²
dA = dx.dy = r.dr.dt

Let's denote I your first double integral :

I = Integral ( t = 0 to pi/4 ; Integral ( r = 0 to sqrt(pi) / cos(t) ; r.sin(r².cos²(t)) ; dr ) ; dt )

I = Integral ( t = 0 to pi/4 ; [ - cos(r².cos²(t)) / (2.cos²(t)) ], 0 -> sqrt(pi) / cos(t) ; dt )

I = Integral ( t = 0 to pi/4 ; 1/cos²(t) ; dt )

I = [ tan (t) ], 0 -> pi/4

I = tan(pi/4) - tan(0) = 1


Let J be your second double integral.

Your ring is the set of points with coordinates pi < r² < 2.pi and 0 < t < 2.pi

J = Integral ( t = 0 to 2.pi ; Integral ( r = sqrt(pi) to sqrt(2.pi) ; r.[sin(r²)]^6 ; dr ) ; dt )

Let K = Integral ( r = sqrt(pi) to sqrt(2.pi) ; r.[sin(r²)]^6 ; dr )

J = Integral ( t = 0 to 2.pi ; K ; dt ) = 2.pi.K

And now the only problem is to find out the simple integral K :

Use Euler's formula to transform sin^6 :

sin(a)^6 = ( -cos(6a) + 6.cos(4a) - 15.cos(2a) + 10) / 32

In turns :

F(r) = ( - sin(6r²) / 12 + 3.sin(4r²) / 4 - 15.sin(2r²) / 4 + 5.r² ) / 32

is an anti-derivative of f(r) = r.[sin(r²)]^6

It appears that every term with sine equals zero when r = sqrt(pi) or sqrt(2.pi), hence : K = (5.2.pi - 5.pi) / 32 = 5.pi / 32

And J = 5.pi² / 16

Feel free to double-check, and to pm if needed.

Good luck !