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 \(x = \phi(u,v), y = \psi(u,v) \)
Then thr double integral is converted into:
\[ \iint_R f(x + y)dxdy = \iint_R f\left[\phi(u,v), \psi(u,v) \right]|J|dudv \]
where,
\[ dxdy = |J|dudv \]
\[ J = \frac{\partial (x,y)}{\partial (u,v)} \]
\[ =\large
\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\
\frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}
\end{vmatrix}
\]
Sample Problem
Evaluate \( \iint_R (x + y)^2dxdy \) 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:
\[ x + y = 0 \longrightarrow (1) \]
\[ x + y = 2 \longrightarrow (2) \]
\[ 3x - 2y = 0 \longrightarrow (3) \]
\[ 3x - 2y = 3 \longrightarrow (4) \]
Finding point of intersection,
From equation (1) \[ x = -y \]
Putting in (3) \[ 3(-y) - 2y = 0 \]
\[ \boxed{y = 0} \Rightarrow \boxed{x = 0} \]
One point is (0,0)
Putting \(x = -y \) in (4)
\[ 3(-y) - 2y = 3 \]
\[ \Rightarrow \boxed{y = \frac{-3}{5}} \Rightarrow \boxed{x = \frac{3}{5}} \]
Second point is (3/5, -3/5)
Using (2)
\[ x + y = 2 \Rightarrow x = 2 - y \]
Putting in (4)
\[ 3(2 - y) - 2y = 3 \]
\[ 6 - 5y = 3 \Rightarrow \boxed{y = \frac{3}{5}} \]
\[ x = 2 - \frac{3}{5} \Rightarrow \boxed{x = \frac{7}{5}} \]
Third point is (7/5, 3/5)
Putting \(x = 2 - y \) in (3)
\[ 3(2 - y) - 2y = 0 \]
\[ 6 - 5y = 0 \]
\[ \Rightarrow \boxed{y = \frac{6}{5}} \]
\[ x = 2 - \frac{6}{5} \Rightarrow \boxed{x = \frac{4}{5}} \]
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.
.jpg)
Now as we know \(J =\Large \frac{1}{J'} \)
\[ J\left(\frac{u,v}{x,y}\right) = \frac{1}{J\left(\large\frac{x,y}{u,v} \right)} \]
\[ J\left(\frac{u,v}{x,y} \right)
= \Large\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}
\]
\[ = \begin{vmatrix}
1 & 1 \\
3 & -2
\end{vmatrix}
\]
\[ = -2 - 3 = -5 \]
\[ J\left(\frac{x,y}{u,v} \right) = \frac{-1}{5} \]
We can clearly see that limits for \(u\) is 0 to 2
And limits for \(v\) is 0 to 3
\[ \iint_R (x + y)^2dxdy = \iint_{R'} u^2|J|dudv \]
\[ = \int^3_0 \int^2_0 u^2\frac{1}{5}dudv \]
\[ = \frac{1}{5}\int^3_0 \left[\frac{u^3}{3} \right]^2_0 dv \]
\[ = \boxed{\frac{8}{5}} \]
