- Thread starter
- #1

- Thread starter jk8985
- Start date

- Thread starter
- #1

- Admin
- #2

- Thread starter
- #3

Can someone let me know if this is correct and if I showed all steps?

Since the the region of integration inside x^2 + y^2 = 1 (and symmetry), convert to polar coordinates:

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

= 8 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) 2r (1 - r^2 cos^2(θ))^(-1/2) dr dθ

= 8 * ∫(θ = 0 to π/2) (-1/cos^2(θ)) 2(1 - r^2 cos^2(θ))^(1/2) {for r = 0 to 1} dθ

= 16 ∫(θ = 0 to π/2) (1/cos^2(θ)) [1 - (1 - cos^2(θ))^(1/2)] dθ

= 16 ∫(θ = 0 to π/2) (1 - sin θ) dθ/cos^2(θ)

= 16 ∫(θ = 0 to π/2) (sec^2(θ) - sec θ tan θ) dθ

= 16 (tan θ - sec θ) {for θ = 0 to π/2}

= 16 (sin θ - 1)/cos θ {for θ = 0 to π/2}

= 16 (0 - (-1)), using L'Hopital's Rule as θ→ π/2-

= 16.