the region r is the area enclosed by the functions y = 2x and y = x² shown below. the region r is the base…

Explanation:

Step1: Find intersection points

Set $2x = x^{2}$, then $x^{2}-2x = 0$, factoring gives $x(x - 2)=0$. So $x = 0$ and $x = 2$. When $x = 0,y = 0$; when $x = 2,y=4$.

Step2: Express $x$ in terms of $y$

For $y = 2x$, we have $x=\frac{y}{2}$; for $y=x^{2}$, we have $x = \sqrt{y}$.

Step3: Determine diameter of semi - circle

The diameter $d$ of the semi - circle perpendicular to the $y$ - axis is $d=\sqrt{y}-\frac{y}{2}$ (right - hand $x$ value minus left - hand $x$ value).

Step4: Find area of semi - circle

The area of a semi - circle is $A=\frac{1}{2}\pi r^{2}$, and since $r=\frac{d}{2}$, we have $A=\frac{\pi}{8}d^{2}=\frac{\pi}{8}(\sqrt{y}-\frac{y}{2})^{2}$.

Step5: Calculate volume using integral

The volume $V$ of the solid with cross - sectional area $A(y)$ from $y = 0$ to $y = 4$ is given by $V=\int_{a}^{b}A(y)dy$. Here $a = 0,b = 4$ and $A(y)=\frac{\pi}{8}(\sqrt{y}-\frac{y}{2})^{2}$, so $V=\frac{\pi}{8}\int_{0}^{4}(\sqrt{y}-\frac{y}{2})^{2}dy$.

Answer:

$\frac{\pi}{8}\int_{0}^{4}(\sqrt{y}-\frac{y}{2})^{2}dy$