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 July 4th, 2017, 03:26 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle y = x + 1 \Rightarrow u - v + 2 = u + v + 1 + 1 \Rightarrow v = 0$ $\displaystyle y = x - 1 \Rightarrow u-v+2=u+v+1-1 \Rightarrow v = 2$ $\displaystyle u-v+2=-u-v-1+1 \Rightarrow 2u = -2 \Rightarrow u = -1$ $\displaystyle y=-x-1 \Rightarrow u-v+2=-u-v-1-1 \Rightarrow 2u = -4 \Rightarrow u = -2$ 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 $\displaystyle \int_{0}^2 \int^{-1}_{-2} -2(2u+3) du dv$. Tags determinant, double, integral, jacobian Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jhenrique Calculus 5 June 30th, 2015 03:45 PM bereprada Calculus 1 March 17th, 2014 06:20 PM maximus101 Calculus 0 March 4th, 2011 01:31 AM alt20 Calculus 1 September 12th, 2009 08:27 AM abcdefg10645 Calculus 0 May 2nd, 2009 03:47 AM

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