Calculate the double integral.
∫∫ y/(x^2+y^2+1) dA, R=[0,1]x[0,1]
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).
To calculate the double integral, we integrate the function f(x, y) = y/(x^2 + y^2 + 1) over the region R = [0, 1] x [0, 1].
We can express the double integral as follows:
∫∫R y/(x^2 + y^2 + 1) dA
To evaluate this integral, we use a change of variables. Let's introduce new variables u and v such that:
u = x^2 + y^2
v = y
Now, we need to find the Jacobian of the transformation, which is given by:
∂(x,y)/∂(u,v) = ∂x/∂u ∂y/∂v - ∂x/∂v ∂y/∂u
Calculating the partial derivatives, we have:
∂x/∂u = (1/2) * (2x)
∂x/∂v = 0
∂y/∂u = (1/2) * (2y)
∂y/∂v = 1
Substituting these values, the Jacobian becomes:
∂(x,y)/∂(u,v) = (1/2) * (2x * 1 - 0 * 2y) = x
We can rewrite the double integral using the new variables:
∫∫R y/(x^2 + y^2 + 1) dA = ∫∫R (v/u) * |∂(x,y)/∂(u,v)| du dv
Since our region R is defined as [0, 1] x [0, 1], the limits of integration become:
0 ≤ u ≤ 1
0 ≤ v ≤ 1
With the change of variables, our integral becomes:
∫∫R (v/u) * x du dv
Now, we integrate with respect to u first. The limits of integration for u are from 0 to 1.
∫∫R (v/u) * x du dv =
∫[0,1] ∫[0,1] (v/u) * x du dv
Integrating with respect to u, we treat v as a constant:
∫[0,1] (v/u) * x du dv =
(v/u) * ∫[0,1] x du dv =
(v/u) * [xu] evaluated from 0 to 1 =
(v/u) * (x - 0) =
(v/u) * x
Now, we integrate with respect to v. The limits of integration for v are from 0 to 1.
∫∫R (v/u) * x du dv =
∫[0,1] (x) * [(v/u)] dv =
x * ∫[0,1] (v/u) dv
Integrating with respect to v, we treat u as a constant:
x * ∫[0,1] (v/u) dv =
(x/u) * ∫[0,1] v dv =
(x/u) * [v^2/2] evaluated from 0 to 1 =
(x/u) * (1/2 - 0) =
(x/u) * (1/2)
Now, we evaluate this expression within the given limits:
(x/u) * (1/2) =
(x/(x^2 + y^2)) * (1/2)
Therefore, the double integral of y/(x^2 + y^2 + 1) over the region R = [0, 1] x [0, 1] is equal to:
∫∫R y/(x^2 + y^2 + 1) dA = (x/(x^2 + y^2)) * (1/2)