Double integral over rectangle

Question

I was very confused about how to answer this question. Please include steps and using U substitution since this was what topic it was going over in the chapter, thank you! This was the question asked.

double integral of (xysin(x^2+y^2))dxdy (sqrt(4pi/3)<x<0), (sqrt(2pi/3)<y<0). Please round to the nearest 4 decimal places.

Answer 1

First let's do an intermediate result of calculating the indefinite integral

∫xsin(x²+y²)dx

Substitute

u = x²+y²

du/dx = 2x

dx = du/2x

∫xsin(x²+y²)dx = ∫xsin(u)du/2x = ∫sin(u)du/2 = -cos(u)/2 = -cos(x²+y²)/2

Therefore the definite integral of ∫xsin(x²+y²)dx over √(4π/3) < x < 0 is

-cos(4π/3 + y²)/2 + cos(y²)/2

Now lets do the given definite integral

∫∫xysin(x²+y²)dxdy =

∫y (∫xsin(x²+y²)dx) dy =

∫y (-cos(4π/3 + y²)/2 + cos(y²)/2) dy =

-∫ycos(4π/3 + y²)dy/2 + ∫ycos(y²)dy/2 =

-sin(4π/3 + y²) + sin(y²) over √(2π/3) < y < 0

The above two integrals were done using the same kind of u-substitution as illustrated earlier.

The result is

-sin(4π/3 + 2π/3) + sin(4π/3) + sin(2π/3) - sin(0) =

-sin(6π/3) - sin(2π - 4π/3) + sin(2π/3) = 0