Change of Variables in Integration Engineering Maths, Btech first year

Change of Variables in Integration

When double integration becomes too complex, we can also try change of variables.
Consider a double integral problem $${\iint }_{R}f(x+y)dxdy$$ which is to be changed into variables $u,v$. The relation between $u,v$ and $x,y$ is $x=\varphi (u,v),y=\psi (u,v)$
Then thr double integral is converted into: $${\iint }_{R}f(x+y)dxdy={\iint }_{R}f[\varphi (u,v),\psi (u,v)]|J|dudv$$ where, $$dxdy=|J|dudv$$ $$J=\frac{\partial (x,y)}{\partial (u,v)}$$ $$=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v}\end{array}\right|$$

Sample Problem

Evaluate ${\iint }_R(x+y{)}^{2}dxdy$ where R is region bounded by $x+y=0,x+y=2,3x-2y=0,3x-2y=3$. Use transformation $u=x+y,v=3x-2y$.

Solution

First we will trace region R by using given equations: $$x+y=0⟶(1)$$ $$x+y=2⟶(2)$$ $$3x-2y=0⟶(3)$$ $$3x-2y=3⟶(4)$$ Finding point of intersection, From equation (1) $$x=-y$$ Putting in (3) $$3(-y)-2y=0$$ $$\boxed{{\displaystyle y=0}}\Rightarrow \boxed{{\displaystyle x=0}}$$ One point is (0,0)
Putting $x=-y$ in (4) $$3(-y)-2y=3$$ $$\Rightarrow \boxed{{\displaystyle y=\frac{-3}{5}}}\Rightarrow \boxed{{\displaystyle x=\frac{3}{5}}}$$ Second point is (3/5, -3/5)
Using (2) $$x+y=2\Rightarrow x=2-y$$ Putting in (4) $$3(2-y)-2y=3$$ $$6-5y=3\Rightarrow \boxed{{\displaystyle y=\frac{3}{5}}}$$ $$x=2-\frac{3}{5}\Rightarrow \boxed{{\displaystyle x=\frac{7}{5}}}$$ Third point is (7/5, 3/5)
Putting $x=2-y$ in (3) $$3(2-y)-2y=0$$ $$6-5y=0$$ $$\Rightarrow \boxed{{\displaystyle y=\frac{6}{5}}}$$ $$x=2-\frac{6}{5}\Rightarrow \boxed{{\displaystyle x=\frac{4}{5}}}$$ Fourth point is (4/5, 6/5).
So all points of intersection are: (0, 0), (3/5, -3/5), (7/5, 3/5) and (4/5, 6/5). The region formed is a parallelogram.

Now using transformation, $$u=x+y$$ $$v=3x-2y$$ So here we will have 4 equations, $$u=0,u=2,v=0,v=3$$ We will trace the curve. The resulting region is a rectangle.

Now as we know $J=\frac{1}{J^{\prime }}$ $$J\left(\frac{u,v}{x,y}\right)=\frac{1}{J\left(\frac{x,y}{u,v}\right)}$$ $$J\left(\frac{u,v}{x,y}\right)=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|$$ $$=\left|\begin{array}{cc}1& 1\\ 3& -2\end{array}\right|$$ $$=-2-3=-5$$ $$J\left(\frac{x,y}{u,v}\right)=\frac{-1}{5}$$ We can clearly see that limits for $u$ is 0 to 2
And limits for $v$ is 0 to 3 $${\iint }_R(x+y{)}^{2}dxdy={\iint }_{R^{\prime }}{u}^2|J|dudv$$ $$={\int }_0^3{\int }_0^2{u}^2\frac{1}{5}dudv$$ $$=\frac{1}{5}{\int }_0^3{\left[\frac{u^3}{3}\right]}_0^{2}dv$$ $$=\boxed{{\displaystyle \frac{8}{5}}}$$