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July 4th, 2017, 03:26 PM   #1
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double integral with Jacobian determinant

Suppose we want to integrate f(x, y) = x + y over the diamond U with boundary lines

y = x + 1, y = x - 1, y = -x + 1 and y = -x - 1

Fill in the three blank spaces below (two bounds and the integrand) to write the integral of f(x, y) = x + y as an integral in u and v:

Using substitution x = u + v + 1 and y = u - v + 2.

The First and third bounding lines above have already been converted into (u, v) bounds and included below. DO NOT EVALUATE this integral. (Hint: You may use the fact that "Jacobian determinant" = -2 for this substitution.)

$\displaystyle
\int_{0}^{?} \int_{?}^{-1} ????? du dv
$

My question is how to use the Jacobian determinant as substitution to find out the inner lower limit?
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July 4th, 2017, 08:56 PM   #2
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$\displaystyle y = x + 1 \Rightarrow u - v + 2 = u + v + 1 + 1 \Rightarrow v = 0$
$\displaystyle y = x - 1 \Rightarrow u-v+2=u+v+1-1 \Rightarrow v = 2$
$\displaystyle u-v+2=-u-v-1+1 \Rightarrow 2u = -2 \Rightarrow u = -1$
$\displaystyle y=-x-1 \Rightarrow u-v+2=-u-v-1-1 \Rightarrow 2u = -4 \Rightarrow u = -2$

So the first missing limit is 2 and the second missing limit is -2.

As for the Jacobian determinant,

$\displaystyle x + y = 2u + 3 \Rightarrow u = \frac{x+y-3}{2}$
$\displaystyle x - y = 2v - 1 \Rightarrow v = \frac{x-y+1}{2}$


$\displaystyle \dfrac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y}\\[1em]
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{vmatrix}=\begin{vmatrix}
\dfrac{\partial (x+y-3)/2}{\partial x} & \dfrac{\partial (x+y-3)/2}{\partial y}\\[1em]
\dfrac{\partial (x-y+1)/2}{\partial x} & \dfrac{\partial (x-y+1)/2}{\partial y} \end{vmatrix}=\begin{vmatrix}
\frac{1}{2} & \frac{1}{2}\\[1em]
\frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2}$

The Jacobian determinant has two definitions; one is the reciprocal of the other, and your textbook appears to use the other.

Finally, $\displaystyle x + y = u + v + 1 + u - v + 2 = 2u+3$.

So the integral is

$\displaystyle \int_{0}^2 \int^{-1}_{-2} -2(2u+3) du dv$.
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