Verified answer

Convert to polar coordinates.

Then, you are integrating over the unit disk r <= 1 with θ in [0, π/4].

(Note that y = x <==> θ = π/4.)

So, the integral transforms to

∫(θ = 0 to π/4) ∫(r = 0 to 1) (r cos θ)^2 * (r dr dθ)

= ∫(θ = 0 to π/4) cos^2(θ) dθ * ∫(r = 0 to 1) r^3 dr

= ∫(θ = 0 to π/4) (1/2)(1 + cos(2θ)) dθ * ∫(r = 0 to 1) r^3 dr

= [(1/2)(θ + sin(2θ)/2) {for θ = 0 to π/4}] * (1/4)

= (1/8)(π/4 - 1/2).

I hope this helps!

Have you learned to integrate with polar coordinates yet??

x=rcosu

y=rsinu

dxdy = rdrdu

0≤ r ≤ 1

π/4≤ u ≤ 3π/4

∫ ∫ r^2Cos^2(u)rdrdu your limits for r and u are above. You have to rewrite Cos^2(u) using a half angle formula.

(∫Cos^2(u)du) (∫r^3dr)

Hope this helps, you can do it in dydx, you just have to break up the integration so that your functions are well, functions. Solve for "y" in your bounds that are defined as the circle. Really this is MUCH easier in polar...