$Calculate $\iint_R (x+y)^2 e^{x-y}dx dy$ on the given region$

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$Calculate $\iint_R (x+y)^2 e^{x-y}dx dy$ on the given region$

Calculate the integral
\[\iint_R (x+y)^2 e^{x-y}dx dy, ∬ R (x+y) 2 e x−y dxdy, \]
where $R$ is the region bounded by $x+y=1$, $x+y=4$, $x-y=-1$ and $x-y=1$.

Multivariable Calculus Calculus Integrals

Emmadavis

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Let $u=x+y$ and $v=x-y$. Then
\[x=\frac{1}{2}(u+v), \ \ \ \ y=\frac{1}{2}(u-v). \]
Hence

\[\frac{\partial (x,y)}{\partial(u,v)}=|det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}|=|\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{pmatrix}|=\frac{1}{2}.\]
Now by change of variable formula we have
\[\iint_{R}(x+y)^2 e^{x-y}dxdy=\frac{1}{2}\int_{1}^{4}\int_{-1}^{1}u^2 e^{v}dvdu=\frac{1}{2} \int_{1}^{4} u^2 du \int_{-1}^{1}e^{v}dv\] \[\frac{1}{2}(\frac{4^3}{3}-\frac{1^3}{3})(e-e^{-1})=\frac{63}{6}(e-e^{-1}),\]
Where we have used the formulas $\int e^{v}dv=e^{v}+c$ and $\int u^2 du=\frac{u^3}{3}+c.$

Erdos

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$Calculate $\iint_R (x+y)^2 e^{x-y}dx dy$ on the given region$

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