Please walk me through how to do this problem (CALC 3) I'm confused what a Unit disc is.

Question

Compute the double integral R (region) of (x^2+y^2)^2dA where R is the unit disc.

I know I have to set up d-theta and dr, but does the bounds of a unit disc (which I'm assuming is 0 to 2pi)? Apply to both dr and d-theta? Can you explain step by step how to set up the integral?

Answer 1

Convert to polar coordinates:

x = rcosθ and y = rsinθ

So, x²+y² = r²cos²θ + r²sin²θ = r²(cos²θ + sin²θ) = r²(1) = r²

On the unit disc, for each r from 0 to 1, θ varies from 0 to 2π.

∫ ∫_R (x²+y²)²dA =

∫_{(0 to 2π)} ∫_{(0 to 1)} (r²)²rdrdθ = ∫_{(0 to 2π)} [(1/6)r⁶_{(0 to 1)}]dθ = (1/6)∫_{(0 to 2π)}dθ = (1/6)θ_{(0 to 2π)} = π/3

Answer 2

We use polar system: x = rcosθ; y = rsinθ; dA = dxdy = rdrdθ; x² + y² = r²; R = {0 ≤ r ≤ 1; 0 ≤ θ ≤ 2π}

∫ ∫_R (x²+y²)²dA = ∫₀¹r⁵dr∫₀^2πdθ = r⁶/6₀¹θ₀^2π = π/3