Change of Variables — Question 1

Question 1 – Solution

Strategy. The boundary expressions themselves become coordinates, turning the tilted rectangle into an axis-aligned rectangle.

Step 1: Transform the region

The inequalities become $-2\le u\le 2$ and $-1\le v\le 1$. Solving $u=x+y$, $v=x-y$ gives $$x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}.$$

Step 2: Jacobian $$\frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}\frac 12&\frac 12\\[2pt]\frac 12&-\frac 12\end{vmatrix} =-\frac 12, \qquad dA=\frac 12\,du\,dv.$$

Step 3: Evaluate $$\begin{align*} I&=\int_{-2}^{2}\int_{-1}^{1}u^2\left(\frac 12\right)dv\,du\\ &=\frac 12(2)\left[\frac{u^3}{3}\right]_{-2}^{2} =\boxed{\frac{16}{3}}. \end{align*}$$

Verification The integrand is nonnegative. The transformed rectangle has area $8$, scaled by $1/2$, so $\operatorname{area}(R)=4$; the result corresponds to average value $4/3$, which is below the maximum $u^2=4$.