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Make appropriate changes of variables in the integral ∬R4(x−y)2dydx∬R4(x−y)2dydx where R is the trapezoid bounded by the lines x - y = 2 , x - y = 4, x = 0, y = 0. Write the resulting integral. My attempt to answer this question: u = x - y, v= x + y. x=12(v+u),y=12(v−u)x=12(v+u),y=12(v−u) $ J ( u, v) = \frac{\partial (x,y)}{\partial(u,v)} =\begin{vmatrix} \frac{\partial{x}}{\partial{u}} & \frac{\partial{x}}{\partial{v}} \end{vmatrix}$ = $\begin{vmatrix}\frac12 & \frac12 \\ -\frac12 & \frac12 \end{vmatrix}= \frac14 - (-\frac14) =\frac12$ Original integrand becomes $\frac{4}{u^2}$ Therefore, using the transformation T, the integral changes to $\displaystyle\iint\limits_R \frac{4}{(x-y)^2} dy dx = \displaystyle\int_?^?\displaystyle\int_?^? \frac{4}{u^2}\frac12 du dv$ Now, how to compute the integration limits?
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the area of the trapezoid is \(\displaystyle 6\) on the other hand the integral \(\displaystyle \displaystyle\iint\limits_R \frac{4}{(x-y)^2} dy dx=\int _2^4\int _{-u}^u\frac{2}{u^2}dvdu=4 \ln 2\) your answer \(\displaystyle \displaystyle\int_2^4 \displaystyle\int_0^{12} \frac{4}{u^2}\frac12 dv du =6\) is incorrect
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Idea said: \(\displaystyle \int _2^4\int _{-u}^u\frac{2}{u^2}dvdu=\log 16\) Click to expand... Would you explain what is $4\ln(2)$? I think we have already answered this problem with alternative answer also.
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\(\displaystyle \int _2^4\int _{-u}^u\frac{2}{u^2}dvdu=\log 16\)
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I rewrite this question again so that readers, viewers, mathematics students can read it very easily. Make appropriate changes of variables in the integral $\displaystyle\iint\limits_R \frac{4}{(x-y)^2} dy dx $ where R is the trapezoid bounded by the lines x - y = 2 , x - y = 4, x = 0, y = 0. Write the resulting integral. We can answer this question, alternatively. $\displaystyle\int_2^4 \displaystyle\int_0^{12} \frac{4}{u^2}\frac12 dv du =6$ Please read #1, #2, #3 for more details
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in the u-v plane, we have area of trapezoid \(\displaystyle \int _2^4\int _{-u}^u\frac{1}{2}dvdu=6\)
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WMDhamnekar said: Make appropriate changes of variables in the integral ∬R4(x−y)2dydx∬R4(x−y)2dydx where R is the trapezoid bounded by the lines x - y = 2 , x - y = 4, x = 0, y = 0. Write the resulting integral. My attempt to answer this question: u = x - y, v= x + y. x=12(v+u),y=12(v−u)x=12(v+u),y=12(v−u) $ J ( u, v) = \frac{\partial (x,y)}{\partial(u,v)} =\begin{vmatrix} \frac{\partial{x}}{\partial{u}} & \frac{\partial{x}}{\partial{v}} \end{vmatrix}$ = $\begin{vmatrix}\frac12 & \frac12 \\ -\frac12 & \frac12 \end{vmatrix}= \frac14 - (-\frac14) =\frac12$ Original integrand becomes $\frac{4}{u^2}$ Therefore, using the transformation T, the integral changes to $\displaystyle\iint\limits_R \frac{4}{(x-y)^2} dy dx = \displaystyle\int_?^?\displaystyle\int_?^? \frac{4}{u^2}\frac12 du dv$ Now, how to compute the integration limits? Click to expand... Now, Putting x = 0, in x - y =2 , we get y = -2, putting y = 0 in x-y =2 , we get x =2. Now, putting x =0 in x-y =4 , we get y =-4 , putting y = 0 in x - y =4 , we get x =4 . So the integration limits are $\displaystyle\int_{-4}^4 \displaystyle\int_{-2}^2 \frac{4}{u^2}\frac12 du dv$ We can observe that the area of this trapezoid is 6. So, how can we solve the integral(computed above) to get the answer 6?
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