The value of the double integral ∬D y dA is -81620 + 192v.
To calculate the double integral ∬D y dA over the region D, we first need to determine the transformation from the region R to the region D using the transformation φ(u, v) = (u², u + v).
The region R is defined as [2, 6] × [0, 2].
Let's find the bounds of the transformed region D under the transformation φ(u, v).
When u varies from 2 to 6 and v varies from 0 to 2, we have:
For u = 2: φ(2, v) = (4, 2 + v)
For u = 6: φ(6, v) = (36, 6 + v)
So, the transformed region D is a region in the uv-plane bounded by u² ranging from 4 to 36 and u + v ranging from 2 to 8.
The integral becomes:
∬D y dA = ∫∫(over D) y dA = ∫(from 4 to 36)∫(from 2 to 8) y dy du
However, since the function y is not explicitly given in terms of u and v, we'll use the inverse transformation to express y in terms of u and v.
The inverse of φ(u, v) = (u², u + v) is given by:
u = u
y = v - u²
Therefore, y = v - u².
Now, we substitute y = v - u² into the integral:
∬D y dA = ∫(from 4 to 36)∫(from 2 to 8) (v - u²) dy du
Now, integrate y with respect to v from 2 to 8:
∬D y dA = ∫(from 4 to 36) [(v * (8 - 2)) - (u² * (8 - 2))] du
∬D y dA = ∫(from 4 to 36) (6v - 6u²) du
Integrating with respect to u from 4 to 36:
∬D y dA = ∫(from 4 to 36) (6(6 + v) - 6u²) du
∬D y dA = ∫(from 4 to 36) (36 + 6v - 6u²) du
Now integrate with respect to u:
∬D y dA = [36u + 6vu - 2u³] evaluated from 4 to 36
∬D y dA = [(36 * 36 + 6 * 36 * v - 2 * 36³) - (36 * 4 + 6 * 4 * v - 2 * 4³)]
∬D y dA = [1296 + 216v - 82944 - 144 - 24v + 128]
∬D y dA = [-81620 + 192v]
So, the value of the double integral ∬D y dA is -81620 + 192v.