2. find the volume of the solid obtained by rotating the region bounded by (y = 1 + x^{2}), (x = 1), (x = 4)…

Explanation:

Step1: Recall the disk - method formula

The formula for the volume $V$ of the solid of revolution about the $x$-axis using the disk - method is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$, where $y = f(x)$ is the function and $[a,b]$ is the interval of integration. Here, $f(x)=1 + x^{2}$, $a = 1$, and $b = 4$.

Step2: Expand the integrand

$(1 + x^{2})^{2}=1 + 2x^{2}+x^{4}$. So the integral becomes $V=\pi\int_{1}^{4}(1 + 2x^{2}+x^{4})dx$.

Step3: Integrate term - by - term

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

Step4: Evaluate the definite integral

$V=\pi\left[\left(x+\frac{2}{3}x^{3}+\frac{1}{5}x^{5}\right)\big|_{1}^{4}\right]=\pi\left[\left(4+\frac{2}{3}(4)^{3}+\frac{1}{5}(4)^{5}\right)-\left(1+\frac{2}{3}(1)^{3}+\frac{1}{5}(1)^{5}\right)\right]$. First, calculate the values inside the brackets: $\left(4+\frac{2}{3}(64)+\frac{1}{5}(1024)\right)-\left(1+\frac{2}{3}+\frac{1}{5}\right)$ $=\left(4+\frac{128}{3}+\frac{1024}{5}\right)-\left(\frac{15 + 10+3}{15}\right)$ $=\left(\frac{60 + 640+3072}{15}\right)-\left(\frac{28}{15}\right)$ $=\frac{60 + 640+3072-28}{15}=\frac{3744}{15}$. So $V=\frac{3744\pi}{15}=\frac{1248\pi}{5}$.

Answer:

$\frac{1248\pi}{5}$