Evaluate: $\int_0^2\int_0^{2-x}(x+y)^2 e^{2y\over x+y}dy dx$.

Question

Evaluate: $$ \int_0^2\int_0^{2-x}(x+y)^2 e^{2y\over x+y}dy dx. $$

My attempt: I've changed the order and got this as $$ \int_{y=0}^2\int_{x=0}^{2-y}(x+y)^2 e^{2y\over x+y}dy dx. $$ and then substituted ${2y\over x+y}=t$ so the integral becomes $$ 8\int_{y=0}^2\int_{t=y}^{2}e^t t^{-4}\,dt\,y^3 dy $$ from here it's becoming very absurd. Please help me to solve easily from here onward, any other short cut will be very benefited. Thanks in advance.

Answer 1

By using the change of variables $$ u = y+x, \\ v = y-x $$ with inverse transformation $$ x = \frac{u-v}{2}, \\ y = \frac{u+v}{2} $$ then the domain can be expressed as $$ 0\leq u\leq 2, \quad -u\leq v\leq u $$ Taking into account that the Jacobian of the transformation is $J=1/2$, we have $$ \frac{1}{2}\int_0^2 du\int_{-u}^{+u} u^2\exp\left(\frac{u-v}{u}\right)dv = 2(e^2-1) $$

Answer 2

you can just take $t=x+y$ it becomes: $$ \int_0^2 \int_y^2 t^2 e^{\frac{2y}{t}}dtdy $$ change the order of integral : $$ \int_0^2 t^2 \int_0^t e^{\frac{2y}{t}}dy dt = \int_0^t t^2(\frac{t}{2})(e^2-1)dt=(e^2-1)\frac{2^4}{2\times 4} = 2(e^2-1) $$