Evaluate double integral √(y^2+z^2) over the region R in the Y-Z plane bounded by y^2+z^2=4.

Answer 1

I = ∫∫_R√y2 + z²dydz; R = { x² + z² ≤ 4); We sweech to polar coordinates: y = rsinθ; z = rcosθ

√y2 + z² = r; dydz = rdrdθ; R= { r ≤ 2, 0 ≤ θ ≤ 2π_R }

I = ∫∫r²drdθ = ∫₀²r²dr∫₀^2πdθ = r³/3₀²·θ₀^2π = 16π/3;