Solve the problem.
-Find the volume of the solid that is generated by revolving the area bounded by the $x$-axis, the curve $y = \sqrt { \frac { 5 x } { x ^ { 2 } + 1 } } , x = 1$, and $x = 3$ about the $x$-axis.
A) $\frac { 5 } { 4 } \pi \ln \left( \frac { 1 } { 5 } \right)$
B) $\frac { 5 } { 2 } \pi \ln ( 5 )$
C) $\frac { 5 } { 2 } \ln ( 5 )$
D) $5 \pi \ln \left( \frac { 1 } { 5 } \right)$

Question

Quizplus · Accepted Answer

First, we need to set up the integral to find the volume. Since we're revolving around the x-axis, we're using the formula $\pi\int_{a}^{b} [f(x)]^{2} dx$, where $f(x)$ is the function defining our solid.
Our biggest challenge is to get the function in terms of x without a square root. So, we make a substitution $u = x^{2} + 1$. Then $du/dx = 2x$ and $x\,dx = du/ 2$. Making these substitutions, we can simplify the function and write the integral:
\begin{align*}
\pi\int_{1}^{3} \frac{5x}{x^{2}+1}dx &= 5\pi\int_{2}^{10} \frac{1}{2(u-1)} du \
&= 5\pi\ln{\mid u - 1\mid}\Big|_{2}^{10} \
&= 5\pi\ln{(\frac{9}{1})} \
&= \frac{5}{2}\pi\ln{(5)} \
\end{align*}
Therefore, the answer is $\boxed{	extbf{(B)}}$.

Solve the Problem xxx -Axis, the Curve y=5xx2+1,x=1y = \sqrt { \frac { 5 x } { x ^ { 2 } + 1 } } , x = 1y=x2+15x​​,x=1

Solve the Problem $x$ -Axis, the Curve $y = \sqrt { \frac { 5 x } { x ^ { 2 } + 1 } } , x = 1$