Answer:
To evaluate the integral over the region R, we can use the change of variables:
u = 4x
v = 2y
This gives us:
x = u/4
y = v/2
The Jacobian of this transformation is:
| ∂x/∂u ∂x/∂v | = | 1/4 0 |
| ∂y/∂u ∂y/∂v | | 0 1/2 |
So the Jacobian determinant is |J| = (1/4)(1/2) = 1/8.
Using this transformation, the region R is mapped onto the unit circle in the uv-plane, and the equation of the ellipse becomes:
u^2 + v^2/4 = 1/16
The integral becomes:
∫∫R 4x^2 e^(4xy) dA
= 2∫∫S u^2 e^uv/2 (1/8) dA
= (1/4) ∫∫S u^2 e^v/2 dA
where S is the unit circle in the uv-plane.
Now we can use polar coordinates in the uv-plane, with u = r cosθ and v = r sinθ. The integral becomes:
(1/4) ∫∫S r^2 cos^2θ e^(r sinθ/2) r dr dθ
= (1/4) ∫0^2π ∫0^1 r^3 cos^2θ e^(r sinθ/2) dr dθ
The inner integral can be evaluated by integration by parts, letting u = r^2 cos^2θ and dv = e^(r sinθ/2) r dr. This gives:
∫ r^3 cos^2θ e^(r sinθ/2) dr
= r^3 cos^2θ (-2/θ) e^(r sinθ/2) + 2/θ ∫ r^2 cos^2θ e^(r sinθ/2) dr
The integral on the right-hand side can be evaluated by another integration by parts, letting u = r^2 cos^2θ and dv = e^(r sinθ/2) dr, which gives:
∫ r^2 cos^2θ e^(r sinθ/2) dr
= r^2 cos^2θ (-2/θ) e^(r sinθ/2) + 4/θ^2 ∫ r cos^2θ e^(r sinθ/2) dr
We can substitute these results back into the original integral and simplify to get:
∫∫R 4x^2 e^(4xy) dA
= (1/4) ∫0^2π ∫0^1 r^3 cos^2θ e^(r sinθ/2) dr dθ
= (1/2π) ∫0^π ∫0^1 r^3 cos^2θ e^(r sinθ/2) dr dθ
Now we can evaluate the inner integral:
∫0^1 r^3 cos^2θ e^(r sinθ/2) dr = (1/2) ∫0^1 r^2 e^(r sinθ/2) d(r^2)
= (1/2) ∫0^1 u^(1/2) e^(u sinθ/2) du
Letting t = u sin(θ/2) and using the identity sin(θ/2) = 2