Calculus Double integral

Question

Find ∫∫R−x+2y−4x+5ydA, where R is the parallelogram enclosed by the lines −x+2y=0, −x+2y=9, −4x+5y=1, −4x+5y=4

Answer 1

WE change variables: - x + 2y = u, - 4x + 5y = v. Then R: u = 0, u = 9, v = 1 and v = 4 (rectangle).

y = (4u - v)/3, x = 2y - u = (5u - 2v)/3;

Jacobian J = I5/3 -2/3I

I4/3 -1/3I = 1/3

∫∫_R(- x + 2y - 4x + 5y)dA = ∫₀⁹du∫₁⁴1/3(u + v)dv = 1/3∫₀⁹(uv + v²/2)₁⁴du = 1/3·∫₀⁹(3u + 15/2)du = 1/3(3u²/2 + 15/2u)₀⁹ = 1/3(243/2 + 135/2) = 63

Answer 2

Let us find the the region on the xy-plane that those two pairs of parallel lines define.

With other words let us find the coordinates of the vertices of the parallelogram.

1. The lines -x+2y=0 and -4x+5y = 4 intersect at the point A (-8/3, -4/3)
2. The lines -x+2y=0 and -4x+5y = 1 intersect at the point B (-2/3, -1/3)
3. The lines -x+2y=9 and -4x+5y = 1 intersect at the point C (43/3, 35/3)
4. The lines -x+2y=9 and -4x+5y = 4 intersect at the point D (37/3, 32/3)

Now let u= -x+2y and v= -4x+5y.

Then we can obtain the four points in the uv-plane

They are as follows A' (0, 4), B' (0,1), C' ( 9,1), D' (9,4).. This is telling us that

in the uv-plane the region of integration Ω is a nice rectangle [0,9] χ [1,4]

Solving the above system for x and y we get

x = (5u -2v)/3 and y = (4u-v)/3

Then the Jacobian of the transformation is as follows∂

∂x/∂u = 5/3 ∂x/∂v = -2/3∂

∂y∂u = 4/3 ∂y/∂v = -1/3

Hence | ∂(x,y)/∂(u,v) | = 1/3

Then ∫∫_R(-x+2y-4x+5y) dA = ∫∫_Ω( u+v) ⋅1/3 dv du = 1/3 ∫^u=9_u=0 ∫^v=4_v=1 [u+v] dv du =

1/3 ∫⁹₀ [uv + v²/2 ] |^v=4_v=1 du = 1/3 ∫⁹₀ [4u -17/2] du = 57/2