Evaluate �è R xy dA , where R is the region bounded by x^2 +y^2 <=16 , x>=0 and y>=0
To evaluate the given integral, we need to find the boundaries of the region and express the integral in terms of these boundaries.
The region R is defined by three conditions:
1. x^2 + y^2 <= 16 (this represents the disk centered at the origin with radius 4)
2. x >= 0 (since x is greater than or equal to zero, the region is restricted to the right half of the disk)
3. y >= 0 (similarly, y is greater than or equal to zero, so the region is further restricted to the upper half of the disk)
Combining these conditions, we can express the region R more explicitly as:
R = {(x, y) | 0 <= x <= 4, 0 <= y <= √(16-x^2)}
Now let's evaluate the integral:
∫∫R xy dA
We will integrate first with respect to y and then with respect to x.
∫∫R xy dA = ∫[0 to 4] ∫[0 to √(16-x^2)] xy dy dx
First, let's integrate with respect to y:
∫[0 to √(16-x^2)] xy dy = x * ∫[0 to √(16-x^2)] y dy
Integrating y with respect to y, we get:
x * [y^2/2] evaluated from 0 to √(16-x^2) = x * (16 - x^2)/2
Now, we can integrate this expression with respect to x:
∫[0 to 4] x * (16 - x^2)/2 dx
Expanding and simplifying, we get:
1/2 * ∫[0 to 4] (16x - x^3) dx
Integrating term by term, we have:
1/2 * [(16x^2/2) - (x^4/4)] evaluated from 0 to 4
Simplifying further, we get:
1/2 * [(8x^2) - (x^4/4)] evaluated from 0 to 4
Now we substitute the bounds:
1/2 * [(8(4)^2) - ((4)^4/4)] - 1/2 * [(8(0)^2) - ((0)^4/4)]
Simplifying:
1/2 * [128 - 16] - 0 = 1/2 * 112 = 56
Therefore, ∫∫R xy dA = 56.