Evaluate ∬R xy dA, where R is the region bounded by x^2+y^2≤36, x≥0 and y≥0.

To evaluate the double integral ∬R xy dA, where R is the region bounded by x^2 + y^2 ≤ 36, x ≥ 0, and y ≥ 0, we'll use polar coordinates.

In polar coordinates, we have x = r cos(theta) and y = r sin(theta), where r is the distance from the origin and theta is the angle measured from the positive x-axis.

First, let's determine the limits of integration for r and theta.

Since x^2 + y^2 ≤ 36, we have r^2 ≤ 36. Taking the square root of both sides, we get r ≤ 6.

Since x ≥ 0 and y ≥ 0, we have 0 ≤ theta ≤ pi/2.

Now, let's convert the integral to polar coordinates.

The integrand xy becomes r*cos(theta)*r*sin(theta) = r^2*cos(theta)*sin(theta).

The differential area element dA in polar coordinates is r dr d(theta).

Therefore, the double integral becomes:

∬R xy dA = ∫(0 to pi/2) ∫(0 to 6) r^2*cos(theta)*sin(theta) * r dr d(theta)

We can separate the variables and integrate with respect to r first:

∫(0 to 6) r^3*cos(theta)*sin(theta) dr = (1/4)*6^4*cos(theta)*sin(theta) = 54*cos(theta)*sin(theta)

Now, we can integrate with respect to theta:

∫(0 to pi/2) 54*cos(theta)*sin(theta) d(theta) = -27*cos^2(theta) evaluated from 0 to pi/2 = -27*[cos^2(pi/2) - cos^2(0)] = -27*[0 - 1] = 27

Therefore, the value of the double integral ∬R xy dA is 27.

To evaluate the double integral ∬R xy dA, we need to find the limits of integration for both x and y.

First, let's visualize the given region R. We are given that R is bounded by the equation x^2 + y^2 ≤ 36, x ≥ 0, and y ≥ 0. This represents the upper half of the circle centered at the origin with a radius of 6.

To determine the limits of integration for x, we can see that x varies from 0 to the value where the circle intersects the y-axis. For this circle, the intersection point on the y-axis occurs when x = 0, y = 6. Therefore, the upper limit for x is 0, and the lower limit is the equation of the circle, x = ±√(36 - y^2), but since we are looking in the first quadrant, we will only use the positive value: x = √(36 - y^2).

To determine the limits of integration for y, we can see that y varies from 0 to the value where the circle intersects the x-axis. For this circle, the intersection point on the x-axis occurs when x = 6, y = 0. Therefore, the upper limit for y is 0, and the lower limit is y = 0.

Now we can set up and evaluate the double integral:

∬R xy dA = ∫[0 to 6]∫[0 to √(36 - y^2)] xy dx dy

To evaluate this integral, we integrate with respect to x first, and then with respect to y. Let's perform the integration:

∫[0 to √(36 - y^2)] xy dx = [(1/2)yx^2] from x = 0 to x = √(36 - y^2)
= (1/2)y(√(36 - y^2))^2 - (1/2)y(0)^2
= (1/2)y(36 - y^2)

Now, substitute this result back into the original integral:

∬R xy dA = ∫[0 to 6] [(1/2)y(36 - y^2)] dy

Now we integrate with respect to y:

(1/2) ∫[0 to 6] [36y - y^3] dy
= (1/2)[(18y^2 - (1/4)y^4)] from y = 0 to y = 6
= (1/2)[(18(6)^2 - (1/4)(6)^4) - (18(0)^2 - (1/4)(0)^4)]
= (1/2)[(18(36) - (1/4)(1296) - 0]
= (1/2)(648 - 324)
= (1/2)(324)
= 162

Therefore, the value of the double integral ∬R xy dA is 162.

∫[0,6] ∫[0,√(36-x^2)] xy dy dx

∫[0,6] 1/2 xy^2 [0,√(36-x^2)] dx
∫[0,6] 1/2 x (36-x^2) dx
9x^2 - 1/8 x^4 [0,6]
162