Let's make the following substitutions:
u = x + y
v = x - y
Then, we have:
x = (u + v) / 2
y = (u - v) / 2
Note that the Jacobian of this transformation is:
|du/dx du/dy| = |1 1|
|dv/dx dv/dy| |1 -1|
and its determinant is 2, which is always positive.
Now, let's find the new bounds of integration.
At (π, 0), we have u = π and v = x - y = π, so v + u = 2π and u - v = 0, which gives:
π ≤ u ≤ 2π
π - u ≤ v ≤ π + u
At (2π, π), we have u = x + y = 3π and v = x - y = π, so v + u = 4π and u - v = 2π, which gives:
2π ≤ u ≤ 4π
u - 2π ≤ v ≤ 2π - u
At (π, 2π), we have u = x + y = 3π and v = x - y = -π, so v + u = 2π and u - v = 4π, which gives:
π ≤ u ≤ 2π
u - 2π ≤ v ≤ 2π - u
At (0, π), we have u = x + y = π and v = x - y = -π, so v + u = 0 and u - v = 2π, which gives:
0 ≤ u ≤ π
u - 2π ≤ v ≤ 2π - u
Now, let's rewrite the integrand using the new variables:
(x - y)^2 * cos^2(x + y) = (v^2 / 4) * cos^2(u)
Then, we have:
∬(x-y)^2 * cos^2(x+y) dA = ∬(v^2 / 4) * cos^2(u) * |J| dudv
= ∫π^(2π) ∫(π - u)^(π + u) (v^2 / 4) * cos^2(u) * 2 dvdu + ∫2π^(4π) ∫(u - 2π)^(2π - u) (v^2 / 4) * cos^2(u) * 2 dvdu + ∫π^0 ∫(u - 2π)^(2π - u) (v^2 / 4) * cos^2(u) * 2 dvdu + ∫0^(π) ∫(u - 2π)^(2π - u) (v^2 / 4) * cos^2(u) * 2 dvdu
Simplifying, we obtain:
= ∫π^(2π) ∫(π - u)^(π + u) (v^2 / 2) * cos^2(u) dvdu + ∫2π^(4π) ∫(u - 2π)^(2π - u) (v^2 / 2) * cos^2(u) dvdu + ∫π^0 ∫(u - 2π)^(2π - u) (v^2 / 2) * cos^2(u) dvdu - ∫0^(π) ∫(u - 2π)^(2π - u) (v^2 / 2) * cos^2(u) dvdu
Note that the last integral has a negative sign because we need to reverse the order of integration to have u as the outer integral.
Using the symmetry of the integrand, we can rewrite this as:
= 4∫π^(2π) ∫0^(π + u) (v^2 / 2) * cos^2(u) dvdu + 4∫π^0 ∫0^(u - π) (v^2 / 2) * cos^2(u) dvdu
= 4 * [2/3 * sin^3(u) |_π^(2π)] + 4 * [2/3 * sin^3(u) |^0_π]
= 32/3
Therefore, the value of the double integral is 32/3.
Use change of variables to evaluate the double integral (x-y)^2 * cos^2(x+y) bounded by the points (pi,0), (2pi,pi), (pi,2pi), (0,pi)
5 answers
First mistake:
(x - y)^2 * cos^2(x + y) = (v^2 / 4) * cos^2(u)
It should be
(x - y)^2 * cos^2(x + y) = v^2 * cos^2(u)
(x - y)^2 * cos^2(x + y) = (v^2 / 4) * cos^2(u)
It should be
(x - y)^2 * cos^2(x + y) = v^2 * cos^2(u)
You are correct, thank you for catching that mistake.
The correct transformation is:
u = x + y
v = x - y
Then, we have:
x = (u + v)/2
y = (u - v)/2
Now, the integrand can be rewritten as:
(x - y)^2 * cos^2(x + y) = v^2 * cos^2(u)
The Jacobian is:
|du/dx du/dy| = |1 1|
|dv/dx dv/dy| |1 -1|
and its absolute value is 2, which is always positive.
The new bounds of integration are:
At (π, 0), we have u = π and v = x - y = π, so v + u = 2π and u - v = 0, which gives:
π ≤ u ≤ 2π
π - u ≤ v ≤ π + u
At (2π, π), we have u = x + y = 3π and v = x - y = π, so v + u = 4π and u - v = 2π, which gives:
2π ≤ u ≤ 4π
u - 2π ≤ v ≤ 2π - u
At (π, 2π), we have u = x + y = 3π and v = x - y = -π, so v + u = 2π and u - v = 4π, which gives:
π ≤ u ≤ 2π
v - 2π ≤ v ≤ 2π - v
At (0, π), we have u = x + y = π and v = x - y = -π, so v + u = 0 and u - v = 2π, which gives:
0 ≤ u ≤ π
-v ≤ v ≤ u
Now, we can evaluate the integral:
∬(x-y)^2 * cos^2(x+y) dA = ∬v^2 * cos^2(u) * |J| dvdu
= ∫π^(2π) ∫(π-u)^(π+u) v^2cos^2(u) 2 dvdvdu + ∫2π^(4π) ∫(u-2π)^(2π-u) v^2cos^2(u) 2 dvdvdu + ∫π^0 ∫(u-π)^π v^2cos^2(u) 2 dvdvdu + ∫0^π ∫(u-π)^π v^2cos^2(u) 2 dvdvdu
= 2 ∫π^(2π) ∫(π-u)^(π+u) v^2cos^2(u) dvdu + 2 ∫2π^(4π) ∫(u-2π)^(2π-u) v^2cos^2(u) dvdu + 2 ∫π^0 ∫(u-π)^π v^2cos^2(u) dvdu
= 2 ∫0^(π) ∫0^u v^2cos^2(u) dvdvdu + 2 ∫0^(π) ∫0^(-u+2π) v^2cos^2(u) dvdvdu + 2 ∫0^(π) ∫(u-π)^π v^2cos^2(u) dvdvdu
= 2/3 [u^3sin^2(u)+u^3cos^2(u)-6u^2sin(u)+12ucos(u)-12sin(u)+8] |^π_0 + 2/3 [u^3cos^2(u)-6ucos(u)+12sin(u)] |^π_0
= 32/3
Therefore, the value of the double integral is 32/3.
The correct transformation is:
u = x + y
v = x - y
Then, we have:
x = (u + v)/2
y = (u - v)/2
Now, the integrand can be rewritten as:
(x - y)^2 * cos^2(x + y) = v^2 * cos^2(u)
The Jacobian is:
|du/dx du/dy| = |1 1|
|dv/dx dv/dy| |1 -1|
and its absolute value is 2, which is always positive.
The new bounds of integration are:
At (π, 0), we have u = π and v = x - y = π, so v + u = 2π and u - v = 0, which gives:
π ≤ u ≤ 2π
π - u ≤ v ≤ π + u
At (2π, π), we have u = x + y = 3π and v = x - y = π, so v + u = 4π and u - v = 2π, which gives:
2π ≤ u ≤ 4π
u - 2π ≤ v ≤ 2π - u
At (π, 2π), we have u = x + y = 3π and v = x - y = -π, so v + u = 2π and u - v = 4π, which gives:
π ≤ u ≤ 2π
v - 2π ≤ v ≤ 2π - v
At (0, π), we have u = x + y = π and v = x - y = -π, so v + u = 0 and u - v = 2π, which gives:
0 ≤ u ≤ π
-v ≤ v ≤ u
Now, we can evaluate the integral:
∬(x-y)^2 * cos^2(x+y) dA = ∬v^2 * cos^2(u) * |J| dvdu
= ∫π^(2π) ∫(π-u)^(π+u) v^2cos^2(u) 2 dvdvdu + ∫2π^(4π) ∫(u-2π)^(2π-u) v^2cos^2(u) 2 dvdvdu + ∫π^0 ∫(u-π)^π v^2cos^2(u) 2 dvdvdu + ∫0^π ∫(u-π)^π v^2cos^2(u) 2 dvdvdu
= 2 ∫π^(2π) ∫(π-u)^(π+u) v^2cos^2(u) dvdu + 2 ∫2π^(4π) ∫(u-2π)^(2π-u) v^2cos^2(u) dvdu + 2 ∫π^0 ∫(u-π)^π v^2cos^2(u) dvdu
= 2 ∫0^(π) ∫0^u v^2cos^2(u) dvdvdu + 2 ∫0^(π) ∫0^(-u+2π) v^2cos^2(u) dvdvdu + 2 ∫0^(π) ∫(u-π)^π v^2cos^2(u) dvdvdu
= 2/3 [u^3sin^2(u)+u^3cos^2(u)-6u^2sin(u)+12ucos(u)-12sin(u)+8] |^π_0 + 2/3 [u^3cos^2(u)-6ucos(u)+12sin(u)] |^π_0
= 32/3
Therefore, the value of the double integral is 32/3.
Second mistake:
|1 1|
| 1 -1| = -2
|1 1|
| 1 -1| = -2
You are correct, thank you for pointing that out. The absolute value of the determinant of the Jacobian should be 2, not -2. The transformation is orientation-preserving, so the Jacobian should be positive. Therefore, we have:
|1 1|
| 1 -1| = 2
My apologies for the confusion.
|1 1|
| 1 -1| = 2
My apologies for the confusion.