Evaluate the double integral ∫∫ y dA over D, where
D is the top half of the disk with center the origin and radius 5?
I guess that would just be
∫[-5,5]∫[0,√(25-x^2)] y dy dx = 250/3
To evaluate the double integral ∫∫ y dA over the region D, which is the top half of the disk with center the origin and radius 5, we will use polar coordinates.
In polar coordinates, the top half of the disk with center at the origin and radius 5 can be represented as the region D: 0 ≤ θ ≤ π and 0 ≤ r ≤ 5.
The integral becomes:
∫∫ y dA = ∫∫ρsin(θ) dA
First, let's evaluate the inner integral with respect to ρ:
∫ρsin(θ) dρ from 0 to 5
= [1/2 ρ^2 sin(θ)] from 0 to 5
= 1/2 (5^2) sin(θ) - 1/2 (0^2) sin(θ)
= 25/2 sin(θ)
Now, let's evaluate the outer integral with respect to θ:
∫ (25/2 sin(θ)) dθ from 0 to π
= [-25/2 cos(θ)] from 0 to π
= -25/2 cos(π) - (-25/2 cos(0))
= -25/2 (-1) - (-25/2)(1)
= -25 + 25/2
= -25/2
Therefore, the value of the double integral ∫∫ y dA over the region D is -25/2.
To evaluate the double integral ∫∫ y dA over D, where D is the top half of the disk with center at the origin and radius 5, we can use the polar coordinates.
First, let's set up the double integral in terms of polar coordinates. In polar coordinates, the region D can be described by the inequalities 0 ≤ r ≤ 5 and 0 ≤ θ ≤ π, where r is the radial distance from the origin and θ is the angle measured from the positive x-axis.
The differential element dA in polar coordinates is given by dA = r dr dθ. Now we can rewrite the double integral:
∫∫ y dA = ∫∫ y r dr dθ
Next, we need to determine the limits of integration for r and θ. Since we are integrating over the top half of the disk, the angle θ varies from 0 to π. The radial distance r varies from 0 to the radius of the disk, which is 5.
Now we can evaluate the double integral:
∫∫ y r dr dθ = ∫[0,π]∫[0,5] y r dr dθ
To evaluate this integral, we need to compute the integral of y with respect to r and then the integral of the resulting expression with respect to θ.
Since y does not depend on r, the integral of y with respect to r is simply y times the integral of r dr over its limits of integration, which is 1/2 r^2 evaluated from 0 to 5:
∫[0,π]∫[0,5] y r dr dθ = ∫[0,π] y (1/2 r^2) evaluated from 0 to 5 dθ
= ∫[0,π] y (1/2 (5)^2 - 1/2 (0)^2) dθ
= ∫[0,π] (25/2) y dθ
Finally, we need to integrate (25/2) y with respect to θ over its limits of integration, which are 0 and π.
∫[0,π] (25/2) y dθ = (25/2) y θ evaluated from 0 to π
= (25/2) y (π - 0)
= (25/2) π y
So, the value of the double integral ∫∫ y dA over D is (25/2) π y.