Question

In this problem we use the change of variables x=5s+tx=5s+t, y=s−ty=s−t to compute the integral ∫r(x+y)da∫r(x+y)da, where rr is the parallelogram with vertices (x,y)=(0,0)(x,y)=(0,0), (5,1)(5,1), (6,0)(6,0), and (1,−1)(1,−1).

Answer 1

$x=5s+t$

$y=s-t$

The Jacobian matrix for this transformation is

$\mathbf J=\dfrac{\partial(x,y)}{\partial(s,t)}=\begin{bmatrix}\dfrac{\partial x}{\partial s}&\dfrac{\partial x}{\partial t}\\\\\dfrac{\partial y}{\partial s}&\dfrac{\partial y}{\partial t}\end{bmatrix}=\begin{bmatrix}5&1\\1&-1\end{bmatrix}\implies\det\mathbf J=-6$

The vertices in the $x,y$ plane correspond to the following points in the $s,t$ plane:

$(x,y)=(0,0)\implies\begin{cases}5s+t=0\\s-t=0\end{cases}\implies(s,t)=(0,0)$

$(x,y)=(5,1)\implies\begin{cases}5s+t=5\\s-t=1\end{cases}\implies(s,t)=(1,0)$

$(x,y)=(6,0)\implies\begin{cases}5s+t=6\\s-t=0\end{cases}\implies(s,t)=(1,1)$

$(x,y)=(1,-1)\implies\begin{cases}5s+t=1\\s-t=-1\end{cases}\implies(s,t)=(0,1)$

That is, the parallelogram $\mathcal R$ is mapped to a square $\mathcal R^*$ in the $s,t$ plane, so

$\displaystyle\iint_{\mathcal R}x+y\,\mathrm dA=\iint_{\mathcal R^*}((5s+t)+(s-t))|\det\mathbf J|\,\mathrm ds,\,\mathrm dt$

$=\displaystyle6\int_{t=0}^{t=1}\int_{s=0}^{s=1}6s\,\mathrm ds\,\mathrm dt=18$