Calculate the double integral ∫∫ xy^2/ x^2+1 dA where the region of integration R= {(x,y)| 0 ≤ x ≤1, -3 ≤ y ≤ 3}

Answer 1

9 ln(2)

integrate y first

int(y**2) = y**3/3

plug in -3 to 3 and you get

54/3 (int(x/(x**2+1)))

use substitution

u=x**2+1

you get 54/6 int(du/u)

54/6 ( ln(x**2+1), x=0..1)

ln(1) = 0 so, 54/6 (ln2(2)) = 9ln(2)

Answer 2

∫_{(from -3 to 3)}y²[∫_{(from 0 to 1)}{x/(x²+1)dx}]dy

= ∫_{(from -3 to 3)}y²[(1/2)ln(x²+1)]_{(from 0 to 1)} dy

= (1/2)ln2∫_{(from -3 to 3)}y²dy = (1/2)ln2)[(1/3)y³)]_{(from -3 to 3)}

= (1/6)ln2)(27 - (-27)) = 9ln2