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# Evaluate the integral where R is the first quadrant region?

Double Integral (x^2 -2y^2) dA, where R is the first quadrant region between circle of R 1 and radius 4

I keep getting 100.128 but its wrong

Thanks

### 1 Answer

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- kbLv 77 years agoFavorite Answer
Using polar coordinates:

∫∫ (x^2 - 2y^2) dA

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

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

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

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

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

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

= (1/2)(-θ + (3/2) sin(2θ)) {for θ = 0 to π/2} * (1/4)r^4 {for r = 1 to 4}

= (1/2) * -π/2 * 255/4

= -255π/16.

I hope this helps!

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