let $r$ be the region enclosed by the line $y = 7$ and the curve $y=x^{2}+3$. a solid is generated by…

Explanation:

Step1: Find the radius of the cross - section

The outer radius $R$ from the line $y = 2$ to $y=7$ is $R=7 - 2=5$. The inner radius $r$ from the line $y = 2$ to $y=x^{2}+3$ is $r=(x^{2}+3)-2=x^{2}+1$. The limits of integration are found by setting $x^{2}+3 = 7$, which gives $x^{2}=4$ or $x=-2$ and $x = 2$.

Step2: Use the washer method formula

The volume $V$ of the solid of revolution using the washer method is $V=\pi\int_{a}^{b}(R^{2}-r^{2})dx$, where $a=-2$, $b = 2$, $R = 5$ and $r=x^{2}+1$. So $V=\pi\int_{-2}^{2}(25-(x^{2}+1)^{2})dx$.

Step3: Expand the integrand

Expand $(x^{2}+1)^{2}=x^{4}+2x^{2}+1$. Then the integrand becomes $25-(x^{4}+2x^{2}+1)=24 - x^{4}-2x^{2}$.

Step4: Integrate term - by - term

$\int(24 - x^{4}-2x^{2})dx=24x-\frac{1}{5}x^{5}-\frac{2}{3}x^{3}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[24x-\frac{1}{5}x^{5}-\frac{2}{3}x^{3}\right]_{-2}^{2}$. $V=\pi\left[\left(24\times2-\frac{1}{5}\times2^{5}-\frac{2}{3}\times2^{3}\right)-\left(24\times(-2)-\frac{1}{5}\times(-2)^{5}-\frac{2}{3}\times(-2)^{3}\right)\right]$. $V=\pi\left[\left(48-\frac{32}{5}-\frac{16}{3}\right)-\left(-48+\frac{32}{5}+\frac{16}{3}\right)\right]$. $V=\pi\left(48-\frac{32}{5}-\frac{16}{3}+48-\frac{32}{5}-\frac{16}{3}\right)$. $V=\pi\left(96-\frac{64}{5}-\frac{32}{3}\right)$. $V=\pi\left(\frac{1440 - 192-160}{15}\right)$. $V=\frac{1088\pi}{15}$.

Answer:

$\frac{1088\pi}{15}$