Question

Let R be the region in the first quadrant bounded by the lines y= -2x+4,y= -2x+7,y=x-2 and y=x+1.By making the substitutions u=2x+y,v=x-y and integrating over a suitable region in the u-v plane. Evaluate the integral ∫∫(2x²-xy-y²)dxdy

Answer 1

The sides of parallelogram are $2x+y=4,\ \ 2x+y=7,\ \ x-y=2,\ \ x-y=-1$ .

By making the substitutions $u=2x+y$ , $v=x-y$ , we get the limits on the integral $4\leq u\leq 7$ and $-1\leq v\leq 2$ .

Now, solve for $x$ and $y$ :

$u+v=(2x+y)+(x-y)=3x$ , $\ \ x=\frac{u+v}{3}$

$u-2v=(2x+y)-(2x-y)=3y$ , $\ y=\frac{u-2v}{3}$ .

Compute the Jacobian:

$J(u,v)=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} \tfrac{1}{3} & \tfrac{1}{3} \\ \tfrac{1}{3} & -\tfrac{2}{3} \end{vmatrix}=-\frac{2}{9}-\frac{1}{9}=-\frac{1}{3}$ and $\big|J(u,v)\big|=\frac{1}{3}$

The original integrand becomes $2x^2-xy-y^2=\frac{2(u+v)^2-(u+v)(u-2v)-(u-2v)^2 }{9}=uv$ .

The double integral is $\iint \limits_R(2x^2-xy-y^2)dx\ dy=\int\limits _{-1}^{2}\int\limits_{4}^7uv \cdot \frac{1}{3}\ du \ dv =\frac{1}{3} \int\limits _{-1}^{2}v dv \cdot \int\limits_{4}^7u\ du =\frac{1}{3}\cdot \frac{v^2}{2}\bigg|^2_{-1} \cdot \frac{u^2}{2}\bigg|_4^7=\frac{1}{3}\cdot \frac{3}{2}\cdot \frac{33}{2}=\frac{33}{4}=8.25$

Answer: $\iint \limits_R(2x^2-xy-y^2)dx\ dy=8.25$