Double Integrals over General Regions — Question 4

Question 4

Let $D$ be the region between $y=x^2$ and $y=2-x$.

Tasks

1. Describe $D$ with vertical slices.

2. Reverse the order, explaining why horizontal slices require two integrals.

3. Compute the area in the reversed order and check it in the original order.

Question 4 – Solution

Strategy. Locate the intersection points and the height where the right boundary changes from the parabola to the line.

Step 1: Vertical description Solving $x^2=2-x$ gives $x=-2,1$. Hence $$D=\{(x,y):-2\le x\le 1,\ x^2\le y\le 2-x\}.$$

Step 2: Reverse and split Horizontal slices begin at $x=-\sqrt y$. For $0\le y\le 1$, they end at $x=\sqrt y$; for $1\le y\le 4$, they end on the line $x=2-y$. Thus $$\iint_D1\,dA =\int_0^1\int_{-\sqrt y}^{\sqrt y}1\,dx\,dy +\int_1^4\int_{-\sqrt y}^{2-y}1\,dx\,dy.$$

Step 3: Area $$\begin{align*} A&=\int_0^1 2\sqrt y\,dy+\int_1^4(2-y+\sqrt y)\,dy\\ &=\frac 43+\left[2y-\frac{y^2}{2}+\frac 23y^{3/2}\right]_1^4 =\frac 43+\frac{19}{6}=\boxed{\frac 92}. \end{align*}$$ The original order gives $\int_{-2}^{1}(2-x-x^2)dx=9/2$, confirming the split.