The integral can be written as:
∫ from 0 to 1 ∫ from 0 to 1 of y/(x^2+y^2+1) dydx
Using polar coordinates, we have:
x = rcosθ
y = rsinθ
And the Jacobian is:
|J| = r
The limits of integration become:
0 ≤ r ≤ 1
0 ≤ θ ≤ π/2
The integral becomes:
∫ from 0 to π/2 ∫ from 0 to 1 of (rsinθ)/(r^2+1) r drdθ
The r term in the numerator cancels with the Jacobian. We can solve the inner integral first:
∫ from 0 to 1 of (sinθ)/(r^2+1) dr
= [1/2 ln(r^2+1)] from 0 to 1
= (1/2)ln(2)sinθ
Integrating with respect to θ:
∫ from 0 to π/2 of (1/2)ln(2)sinθ dθ
= [-1/2 ln(cosθ)] from 0 to π/2
= (1/2)ln(2)
Therefore, the double integral is (1/2)ln(2).
Calculate the double integral.
∫∫ y/(x^2+y^2+1) dA, R=[0,1]x[0,1]
1 answer