Evaluate the double integral

Question

Use polar coordinates to evaluate the double integral

Int Dx ² √x² + y² dA

where D is the part of the annulus 1 ≤ x² + y² < =4 that lies in the positive quadrant, x ≥ 0 and y ≥ 0.

Answer 1

In the first quadrant, we have 0<= θ <= π/2

As 1 <= x² + y² <= 4, 1 <= r <= 2

r dr dθ = dy dx

x² √(x² + y²) dA = r² cos²θ r r dr dθ = r⁴ cos²θ

Thus, I[0,π/2] I[1,2] r⁴ cos²θ dr dθ =

I[0,π/2]1/5 r⁵ cos²θ E[1, 2] dθ =

31/5 I[0,π/2] cos²θ dθ =

31/5 I[0,π/2] 1/2 + 1/2 cos2θ dθ =

31/5(1/2θ + 1/4 sin2θ) E[0, π/2] =

31/5 * π/4 =

31π/20