let the region r be the area enclosed by the function f(x)=x² + 1, the horizontal line y = 1 and the…

let the region r be the area enclosed by the function f(x)=x² + 1, the horizontal line y = 1 and the vertical lines x = 0 and x = 2. find the volume of the solid generated when the region r is revolved about the line y = 1. you may use a calculator and round to the nearest thousandth.

let the region r be the area enclosed by the function f(x)=x² + 1, the horizontal line y = 1 and the vertical lines x = 0 and x = 2. find the volume of the solid generated when the region r is revolved about the line y = 1. you may use a calculator and round to the nearest thousandth.

Answer

Explanation:

Step1: Identify the radius function

The distance from the curve $y = x^{2}+1$ to the axis of rotation $y = 1$ is $r(x)=(x^{2}+1 - 1)=x^{2}$. The limits of integration are from $x = 0$ to $x=2$.

Step2: Apply the disk - method formula

The formula for the volume $V$ of a solid of revolution about a horizontal axis using the disk - method is $V=\pi\int_{a}^{b}[r(x)]^{2}dx$. Here, $a = 0$, $b = 2$ and $r(x)=x^{2}$, so $V=\pi\int_{0}^{2}(x^{2})^{2}dx=\pi\int_{0}^{2}x^{4}dx$.

Step3: Integrate the function

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int_{0}^{2}x^{4}dx=\left[\frac{x^{5}}{5}\right]_{0}^{2}=\frac{2^{5}}{5}-\frac{0^{5}}{5}=\frac{32}{5}$.

Step4: Calculate the volume

Multiply by $\pi$: $V=\pi\times\frac{32}{5}=\frac{32\pi}{5}\approx20.106$.

Answer:

$20.106$