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?