Change of Variables — Question 7

Question 7 – Solution

Strategy. Measure vertical displacement from the lower parabola; the curved strip then becomes a rectangle.

Step 1: Straighten the region

The inverse map is $$x=u,\qquad y=u^2+v,$$ and the bounds become $0\le u\le 2$, $0\le v\le 3$.

Step 2: Jacobian and integral $$\frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}1&0\\2u&1\end{vmatrix}=1.$$ Because $y-x^2=v$, $$I=\int_0^2\int_0^3v\,dv\,du =(2)\left[\frac{v^2}{2}\right]_0^3 =\boxed{9}.$$

Verification For each fixed $x$, the displacement $t=y-x^2$ ranges from $0$ to $3$. Thus direct integration gives $\int_0^2\int_0^3t\,dt\,dx=9$. The unit Jacobian reflects a vertical shear with no area scaling.