region $r$ is enclosed by the line $y = 1$ and the curve $y=\frac{1}{4}x^{2}$. what is the volume of the…

Explanation:

Step1: Find intersection points

Set $\frac{1}{4}x^{2}=1$, then $x^{2} = 4$, so $x=-2$ and $x = 2$.

Step2: Use disk - washer method formula

The volume $V$ of the solid of revolution about the $x$-axis using the washer method is $V=\pi\int_{a}^{b}(R^{2}-r^{2})dx$, where $R$ is the outer - radius and $r$ is the inner - radius. Here, $R = 1$ and $r=\frac{1}{4}x^{2}$, and $a=-2$, $b = 2$. So $V=\pi\int_{-2}^{2}(1^{2}-(\frac{1}{4}x^{2})^{2})dx$.

Step3: Simplify the integrand

$V=\pi\int_{-2}^{2}(1 - \frac{1}{16}x^{4})dx$. Since the integrand $y = 1-\frac{1}{16}x^{4}$ is an even function, we can rewrite the integral as $V = 2\pi\int_{0}^{2}(1-\frac{1}{16}x^{4})dx$.

Step4: Integrate term - by - term

$\int(1-\frac{1}{16}x^{4})dx=x-\frac{1}{16}\times\frac{1}{5}x^{5}+C=x-\frac{1}{80}x^{5}+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[x-\frac{1}{80}x^{5}\right]_{0}^{2}=2\pi\left(2-\frac{1}{80}\times32\right)=2\pi\left(2-\frac{2}{5}\right)=2\pi\times\frac{8}{5}=\frac{16\pi}{5}$.

Answer:

$\frac{16\pi}{5}$