Solve the problem.
-Find the volume of the solid that is generated by revolving the area bounded by $y = \frac { 3 } { \sqrt { 8 x + 1 } } , x = 0$, $x = 5$, and $y = 0$ about the $x$-axis.
A) $\frac { 9 } { 8 } \pi \ln ( 41 )$
B) $\frac { 9 } { 8 } \pi \ln ( 8 )$
C) $\frac { 3 \sqrt { 8 } } { 8 } \pi \ln ( 41 )$
D) $\frac { 3 } { 8 } \pi \ln ( 8 )$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the area bounded by the given curves about the x-axis can be found using the disk method. The formula for the volume is $V = \pi \int_{a}^{b} [f(x)]^2 dx$, where $f(x) = \frac{3}{\sqrt{8x+1}}$. Substituting $f(x)$ into the formula and integrating from $x = 0$ to $x = 5$ gives $V = \pi \int_{0}^{5} \left(\frac{3}{\sqrt{8x+1}}\right)^2 dx$, which simplifies to $\frac{9\pi}{8} \int_{0}^{5} \frac{1}{8x+1} dx$. Solving this integral leads to $\frac{9}{8} \pi \ln(41)$, making choice A correct.

Solve the Problem y=38x+1,x=0y = \frac { 3 } { \sqrt { 8 x + 1 } } , x = 0y=8x+1​3​,x=0

Solve the Problem $y = \frac { 3 } { \sqrt { 8 x + 1 } } , x = 0$