Evaluate the following integral by converting to polar coordinate system: ∫∫ye^x dA, where the region R is in the first quadrant enclosed by a circle of radius 5 and centered at origin

Answer 1

x = rcosφ; y = rsinφ;

region R: 0≤r≤5, 0≤φ≤π/2

dA = dxdy = rdrdφ∫

I = ∫∫ye^xdA = ∫∫rsinφe^rcosφrdrdφ = -⌊∫rdre^rcosφ⌊^π/2₀= -∫r(e^rcosπ/2- e^rcos0)dr = ∫r(e^r -1)dr from 0 to 5'

Integration by parts: u = r, du = dr and dv = (e^r -1)dr v = e^r- r

I = r(e^r - r) - ∫(e^r - r)dr = r(e^r - r) - e^r + r²/2 with substitution from 0 to 5

I = 5(e⁵ -5) - e⁵ +25/2 +1 = 4e⁵ -11.5