Question

Use the transformation u=3x+4y,v=x+3y to evaluate the given integral for the region R bounded by the lines y=−43x+2,y=−43x+3,y=−31x, and y=−31x+1

∬R(3x2+13xy+12y2)dx/dy=

Answer 1

To evaluate the given integral using the transformation u=3x+4y,v=x+3y, we need to find the new limits of integration and transform the integrand.

Step-by-step explanation:

To evaluate the given integral using the transformation u=3x+4y,v=x+3y, we need to find the new limits of integration and transform the integrand. Let's start by finding the new limits of integration:

The line y=-43x+2 intersects with the line y=-31x at x=0 and x=2/3.

The line y=-43x+3 intersects with the line y=-31x+1 at x=0 and x=1/4.

The region R is bounded by these intersection points, so the new limits of integration for u and v are:

For u: 0 ≤ u ≤ 3(2/3) + 4(-43)(2/3) = -572/3

For v: 0 ≤ v ≤ (1/4) + 3(-43)(1/4) = -135/4

Now, let's transform the integrand:

3x^2 + 13xy + 12y^2 = (1/9)(u-4v)^2 + (16/9)(v^2)

Finally, we can evaluate the integral over the transformed region R:

∬R (3x^2 + 13xy + 12y^2) dx dy = ∬R ((1/9)(u-4v)^2 + (16/9)(v^2)) du dv