u = x +y
v = -2x +y
∫[R,(blank)] ∫[blank,blank] 2y dx dy
where R is the parallelogram by the lines
y= -x +1
y= -x +4
y= 2x +2
y= 2x +5
Then it wants me to graph the regions R and the transformed region S.
v = -2x +y
∫[R,(blank)] ∫[blank,blank] 2y dx dy
where R is the parallelogram by the lines
y= -x +1
y= -x +4
y= 2x +2
y= 2x +5
Then it wants me to graph the regions R and the transformed region S.
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Your "blank blank" stuff is kind of confusing.
You can plot the four lines given and see that you get a parallelogram. What's more, you can notice that the first two and last two lines can be expressed as
x + y = 1, and x + y = 4,
-2x + y = 2, and -2x + y = 5.
The change of variables u = x + y and v = -2x + y results in a transformed region that is a square in the uv-plane:
1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.
That should similarly be no problem to plot. This also shows that the limits of integration will be 1 to 4 and 2 to 5 for u and v, respectively.
You need the Jacobian. You can use
∂(u, v)/∂(x, y) = |∂u/∂x ∂v/∂y - ∂u/∂y ∂v/∂x| = 3 ==> ∂(x, y)/∂(u, v) = 1/3.
The integrand can be found with a little algebra
2y = (2/3)(2u + v).
So integrate
5 4
∫ ∫ (2/9)(2u + v) du dv = 17
2 1
You can plot the four lines given and see that you get a parallelogram. What's more, you can notice that the first two and last two lines can be expressed as
x + y = 1, and x + y = 4,
-2x + y = 2, and -2x + y = 5.
The change of variables u = x + y and v = -2x + y results in a transformed region that is a square in the uv-plane:
1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.
That should similarly be no problem to plot. This also shows that the limits of integration will be 1 to 4 and 2 to 5 for u and v, respectively.
You need the Jacobian. You can use
∂(u, v)/∂(x, y) = |∂u/∂x ∂v/∂y - ∂u/∂y ∂v/∂x| = 3 ==> ∂(x, y)/∂(u, v) = 1/3.
The integrand can be found with a little algebra
2y = (2/3)(2u + v).
So integrate
5 4
∫ ∫ (2/9)(2u + v) du dv = 17
2 1