Change of Variables in Double Integral (Jacobians)

On this page, we solve few problems on the topic of the change of variables in double integrals using Jacobian method.

Questions- Answers on Change of Variables

$\boxed{\color{blue}\textbf{Q 1}:}$ Evaluate the double integral

I = $\displaystyle \int \displaystyle \int_R x dx dy$

where

R = $\{(x,y) \in \mathbb{R}^2: 1 \leq x(1-y)\leq 2, 1 \leq xy \leq 2\}$.

Answer:

We use the following change of variables.

Put

u = x(1-y)

v = xy

So the above region R can be rewritten as R = $\{(u,v) \in \mathbb{R}^2: 1 \leq u \leq 2, 1 \leq v \leq 2\}$.

Now find the values of x and y in terms of u and v.

We have u+v = x(1-y)+xy = x and y = $\dfrac{v}{x}=\dfrac{v}{u+v}$.

That is,

$\boxed{x=u+v, \quad y=\dfrac{v}{u+v}.}$

Now, Jacobian J(u, v)

= $\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}$

= $\begin{vmatrix} 1 & 1 \\ -\dfrac{v}{(u+v)^2} & \dfrac{(u+v)\cdot 1-v \cdot 1}{(u+v)^2} \end{vmatrix}$

= $\begin{vmatrix} 1 & 1 \\ -\dfrac{v}{(u+v)^2} & \dfrac{u}{(u+v)^2} \end{vmatrix}$

= $\dfrac{u}{(u+v)^2} + \dfrac{v}{(u+v)^2}$

= $\dfrac{u+v}{(u+v)^2}$

= $\dfrac{1}{u+v}$.

So the given double integral

I = $\displaystyle \int \displaystyle \int_R x dx dy$

= $\displaystyle \int_1^2 \displaystyle \int_1^2 \dfrac{1}{u+v} \cdot (u+v) dv \, du$

= $\displaystyle \int_1^2 \displaystyle \int_1^2 dv \, du$

= 1.

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