the region above is revolved around the line x = 3. find the volume accurate to the nearest thousandth…

Explanation:

Step1: Use the method of cylindrical - shells

The formula for the volume $V$ using the method of cylindrical - shells when rotating about a vertical line $x = a$ is $V=2\pi\int_{c}^{d}(a - x)h(x)dx$, where $(a - x)$ is the radius of the shell, $h(x)$ is the height of the shell, and $[c,d]$ is the interval of integration. Here, $a = 3$, $c = 1$, $d = 3$, and $h(x)=x^{2}+3$. So, $V = 2\pi\int_{1}^{3}(3 - x)(x^{2}+3)dx$.

Step2: Expand the integrand

Expand $(3 - x)(x^{2}+3)$: [ \begin{align*} (3 - x)(x^{2}+3)&=3(x^{2}+3)-x(x^{2}+3)\ &=3x^{2}+9 - x^{3}-3x\ &=-x^{3}+3x^{2}-3x + 9 \end{align*} ]

Step3: Integrate term - by - term

[ \begin{align*} \int(-x^{3}+3x^{2}-3x + 9)dx&=-\frac{1}{4}x^{4}+x^{3}-\frac{3}{2}x^{2}+9x+C \end{align*} ]

Step4: Evaluate the definite integral

[ \begin{align*} V&=2\pi\left[-\frac{1}{4}x^{4}+x^{3}-\frac{3}{2}x^{2}+9x\right]_{1}^{3}\ &=2\pi\left[\left(-\frac{1}{4}(3)^{4}+(3)^{3}-\frac{3}{2}(3)^{2}+9(3)\right)-\left(-\frac{1}{4}(1)^{4}+(1)^{3}-\frac{3}{2}(1)^{2}+9(1)\right)\right]\ &=2\pi\left[\left(-\frac{81}{4}+27-\frac{27}{2}+27\right)-\left(-\frac{1}{4}+1-\frac{3}{2}+9\right)\right]\ &=2\pi\left[\left(-\frac{81}{4}+54-\frac{27}{2}\right)-\left(-\frac{1}{4}+10-\frac{3}{2}\right)\right]\ &=2\pi\left[\left(-\frac{81}{4}+\frac{216}{4}-\frac{54}{4}\right)-\left(-\frac{1}{4}+\frac{40}{4}-\frac{6}{4}\right)\right]\ &=2\pi\left[\frac{-81 + 216-54}{4}-\frac{-1 + 40-6}{4}\right]\ &=2\pi\left[\frac{81}{4}-\frac{33}{4}\right]\ &=2\pi\times\frac{48}{4}\ &=24\pi\approx 75.398 \end{align*} ]

Answer:

$75.398$