# Rules for change of variables in double integrals

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.

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.

• Great explanation, thanks! – Ninja Dec 2 '17 at 21:15

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.

• Why this bounds work for $y$? – Ninja Dec 2 '17 at 21:15
• 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 – Believer Dec 2 '17 at 21:22