Double Integrals over General Regions — Question 8

Question 8 – Solution

Strategy. Keep the line slope symbolic; the nonzero intersection becomes the outer endpoint of a vertical-slice area integral.

Step 1: Intersections and bounds From $x^2=kx$, $$x(x-k)=0,$$ so the intersections occur at $x=0,k$. Since $k>0$, $x^2\le kx$ on $[0,k]$, and $$D_k=\{(x,y):0\le x\le k,\ x^2\le y\le kx\}.$$

Step 2: Area formula $$\begin{align*} A(k)&=\int_0^k\int_{x^2}^{kx}1\,dy\,dx =\int_0^k(kx-x^2)dx\\ &=\left[\frac{kx^2}{2}-\frac{x^3}{3}\right]_0^k =\boxed{\frac{k^3}{6}}. \end{align*}$$

Step 3: Inverse condition Set $k^3/6=9/2$. Then $k^3=27$, and the positive restriction gives $$\boxed{k=3}.$$ Substitution yields $A(3)=27/6=9/2$. Since $A(k)=k^3/6$ is strictly increasing for $k>0$, this positive solution is unique.