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July 4th, 2017, 03:26 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 84 Thanks: 1  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? 
July 4th, 2017, 08:56 PM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 852 Thanks: 309 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
$\displaystyle y = x + 1 \Rightarrow u  v + 2 = u + v + 1 + 1 \Rightarrow v = 0$ $\displaystyle y = x  1 \Rightarrow uv+2=u+v+11 \Rightarrow v = 2$ $\displaystyle uv+2=uv1+1 \Rightarrow 2u = 2 \Rightarrow u = 1$ $\displaystyle y=x1 \Rightarrow uv+2=uv11 \Rightarrow 2u = 4 \Rightarrow u = 2$ So the first missing limit is 2 and the second missing limit is 2. As for the Jacobian determinant, $\displaystyle x + y = 2u + 3 \Rightarrow u = \frac{x+y3}{2}$ $\displaystyle x  y = 2v  1 \Rightarrow v = \frac{xy+1}{2}$ $\displaystyle \dfrac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y}\\[1em] \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{vmatrix}=\begin{vmatrix} \dfrac{\partial (x+y3)/2}{\partial x} & \dfrac{\partial (x+y3)/2}{\partial y}\\[1em] \dfrac{\partial (xy+1)/2}{\partial x} & \dfrac{\partial (xy+1)/2}{\partial y} \end{vmatrix}=\begin{vmatrix} \frac{1}{2} & \frac{1}{2}\\[1em] \frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2}$ 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 $\displaystyle \int_{0}^2 \int^{1}_{2} 2(2u+3) du dv$. 

Tags 
determinant, double, integral, jacobian 
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