Question

Evaluate the integral. (4x+8y) dA where R is the parallelogram with vertices (-1,3), (1,-3), (3,-1), and (1,5); 1 x = 1/(u + v); y = - 3u).

Answer 1

To evaluate the integral of (4x+8y) over the parallelogram R, a change of variables is suggested for the integral, requiring transformation mappings, computation of the Jacobian, and double integral computation in the new variable space.

Step-by-step explanation:

The student is asking to evaluate the integral of the function (4x+8y) over a region R, which is defined as a parallelogram with given vertices. To solve this problem, one would typically use a double integral over the region R. However, the question also provides transformations for x and y in terms of u and v, suggesting a change of variables is necessary for the integral.

Firstly, define a suitable coordinate transformation that maps the parallelogram R to a simpler region in u-v space. Then, express x and y in terms of u and v, and compute the Jacobian of the transformation. Lastly, set up and compute the double integral over the new region in u-v space, applying the Jacobian and the transformations for x and y to convert the function (4x+8y) accordingly.

Answer 2

The integral ∫∫_R (4x + 8y) dA, where R is the described parallelogram, evaluates to 0.

Step-by-step explanation:

Change of variables: We are given the transformation:

x = 1/(u + v)

y = -3u

Jacobian determinant: We need to calculate the Jacobian determinant, which represents the area element scaling factor under the transformation:

|J| = |-∂x/∂u ∂x/∂v|

|∂y/∂u ∂y/∂v|

Calculating the partial derivatives:

|J| = |-(u + v)^−2 |u|

|3 |u + v|

Integral substitution: We can substitute the original variables with the transformed ones and multiply by the Jacobian:

∫∫_R (4x + 8y) dA

= ∫∫_S (4(1/(u + v)) + 8(-3u)) |J| du dv

Simplify the integral:

= ∫∫_S (4u^2 - 20uv - 24) |u + v| du dv

Symmetric region and odd integrand: The region R is symmetric across the y-axis, and the integrand (4u^2 - 20uv - 24) is odd in both u and v. When integrating over a symmetric region with an odd function, the positive and negative contributions cancel each other out.

Therefore, the integral ∫∫_R (4x + 8y) dA evaluates to 0.