region $r$ is enclosed by the curve $y = \\frac{1}{2}x^{2}$ and the line $y = 2x$. what is the volume of the…

Explanation:

Step1: Find intersection points

Set $\frac{1}{2}x^{2}=2x$. Rearrange to $x^{2}-4x = 0$, factor out $x$: $x(x - 4)=0$. So $x = 0$ and $x = 4$.

Step2: Use the shell - method formula

The formula for the volume $V$ of the solid of revolution about the $y$-axis using the shell method is $V=2\pi\int_{a}^{b}x\left(f(x)-g(x)\right)dx$, where $a = 0$, $b = 4$, $f(x)=2x$ and $g(x)=\frac{1}{2}x^{2}$.

Step3: Set up the integral

$V = 2\pi\int_{0}^{4}x\left(2x-\frac{1}{2}x^{2}\right)dx=2\pi\int_{0}^{4}\left(2x^{2}-\frac{1}{2}x^{3}\right)dx$.

Step4: Integrate term - by - term

$\int\left(2x^{2}-\frac{1}{2}x^{3}\right)dx=\frac{2}{3}x^{3}-\frac{1}{8}x^{4}+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[\frac{2}{3}x^{3}-\frac{1}{8}x^{4}\right]_{0}^{4}=2\pi\left(\frac{2}{3}(4)^{3}-\frac{1}{8}(4)^{4}\right)$. $V = 2\pi\left(\frac{128}{3}-32\right)=2\pi\left(\frac{128 - 96}{3}\right)=2\pi\times\frac{32}{3}=\frac{64\pi}{3}$.

Answer:

$\frac{64\pi}{3}$