Questions: R is the region bounded by the functions f(x)=1+x^2 and the g(x)=sqrt(x+1) and the lines x=0 and x=3. Represent the volume when R is rotated around the y-axis. Volume =∫ from 0 to 3 square dx

R is the region bounded by the functions f(x)=1+x^2 and the g(x)=sqrt(x+1) and the lines x=0 and x=3.

Represent the volume when R is rotated around the y-axis.
Volume =∫ from 0 to 3 square dx

Transcript text: $R$ is the region bounded by the functions $\mathrm{f}(\mathrm{x})=1+x^{2}$ and the $\mathrm{g}(\mathrm{x})=\sqrt{x+1}$ and the lines $\mathrm{x}=0$ and $\mathrm{x}=3$. Represent the volume when $R$ is rotated around the $y$-axis. Volume $=\int_{0}^{3}$ $\square$ $d x$ Use pi for " $\pi$ " and sqrt(x) for " $\sqrt{x}$ "

Solution

Solution Steps

Step 1: Define the Functions

We are given two functions that bound the region $ R $:

$ f(x) = 1 + x^2 $
$ g(x) = \sqrt{x + 1} $

Step 2: Set Up the Volume Integral

To find the volume of the solid formed by rotating the region $ R $ around the $ y $-axis, we use the method of cylindrical shells. The volume $ V $ is given by the integral: \[ V = \int_{0}^{3} 2\pi x (f(x) - g(x)) \, dx \]

Step 3: Calculate the Volume

Substituting the functions into the integral, we have: \[ V = \int_{0}^{3} 2\pi x \left((1 + x^2) - \sqrt{x + 1}\right) \, dx \] After evaluating this integral, we find: \[ V = \frac{1021\pi}{30} \]