0
$\begingroup$

Consider the integral $I=\displaystyle\int_{R}\int f(x,y)dx dy$ over the region $R$, given by the triangle with vertices $(0,0),(1,1)$ and $(2,0)$.

This is an isosceles triangle with one side lying along the $x-$axis. So, our domain is not "nice" to find the bounds for integral I assume, since even if we write $0\le x \le 2$, we can not give bounds for $y$ easily. To find this integral, the book I am reading makes the following transformation: $u = y-x$, $v=y+x$. After that, our new domain becomes a right angled triangle with the perpendicular edges lying on the $u$ and $v$ axis.

Finally, my question is how can we conclude this transformations? In general, setting up $u = x+y, v= x-y$ works quite nice for triangular/rectangular domains but is there a rule for this?

Thank you.

$\endgroup$
0
$\begingroup$

Considering the case of a triangle. Let the vertices of the triangle be $O$, $P$ and $Q$. We can use a transformation that keeps the vertex $O$ in the same place and takes the vertices $P$ and $Q$ to $P'$ and $Q'$ where $OP'$ and $OQ'$ are aligned parallel to the axes.

We can do this with matrix multiplication if we make a couple of assumptions. Let $ P' = (0, 1) $ and $Q' = (1, 1)$. Let the transforming matrix be $M$, so we have $$ M \begin{pmatrix} P_{x} \\ P_{y} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \textrm{ and } M \begin{pmatrix} Q_{x} \\ Q_{y} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

If we write $$ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ then we can get a system of equations $$ \begin{aligned} P_{x} a + P_{y} b &= 0 \\ P_{x} c + P_{y} d &= 1 \\ Q_{x} a + Q_{y} b &= 1 \\ Q_{x} c + Q_{y} d &= 1 \end{aligned} $$ We can solve for $a$, $b$, $c$ and $d$ in this system to get $M$.

The transformation $$ \begin{pmatrix} u \\ v \end{pmatrix} = M \begin{pmatrix} x \\ y \end{pmatrix} $$ will then make the image $OP'Q'$ of the triangle $OPQ$ in the $(u,v)$ domain one that has sides parallel to the axes and hence easier to integrate over. The determinant of the matrix $M$ would then need to be used as a multiplying factor in the integral. The case for a rectangle should be similar.

$\endgroup$
  • $\begingroup$ Great explanation, thanks! $\endgroup$ – Ninja Dec 2 '17 at 21:15
0
$\begingroup$

By the way region is "nice" if you look this way $\displaystyle\int_{0}^{1}\int_{y}^{2-y}f(x,y)dxdy$

And about how to use change of variables,first try to get all the equations of lines your region have in a format such that right hand side of equations will contain constants only and left hand side will contain some stuff in $x$ and $y$ only.So in this case it was $y-x=0$ and $y+x=2$.So we can put $y-x=u$ and $y+x=v$.So basically we get two lines in $uv$ plane which are $u=0$ and $v=2$.Doing some calculation you will get $u=-v$.So you can construct "nice" $uv$ region easily.

$\endgroup$
  • $\begingroup$ Why this bounds work for $y$? $\endgroup$ – Ninja Dec 2 '17 at 21:15
  • $\begingroup$ Change in order of integration!!Since $dy$ is outside we are concerned only with the min.value and max. value of $y$ which we will easily get by looking at the region in $xy$ plane $\endgroup$ – Believer Dec 2 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.