Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. $y = \ln x , y = 1 , y = 5 , x = 0 ;$ about the $y -$ axis
A) $\frac { \pi } { 4 } \left( e ^ { 10 } - e \right)$
B) $\frac { \pi } { 2 } \left( 1 - e ^ { 10 } \right)$
C) $\frac { \pi } { 2 } \left( e ^ { 10 } - e ^ { 2 } \right)$
D) $\frac { \pi e ^ { 10 } } { 2 }$
E) $2 \pi \left( e ^ { 10 } - e ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The volume of the solid obtained by rotating the region bounded by the given curves about the $y -$ axis is given by the integral $$V = \pi \int_a^b [f(x)]^2 dx$$where $f(x)$ is the function that defines the upper boundary of the region and $a$ and $b$ are the $x$-coordinates of the points where the region intersects the $y$-axis.In this case, $f(x) = \ln x$, $a = 0$, and $b = 1$. Therefore, the volume of the solid is given by the integral$$V = \pi \int_0^1 (\ln x)^2 dx$$This integral can be evaluated using integration by parts. Let $u = (\ln x)^2$ and $dv = dx$. Then $du = 2\ln x \frac{1}{x} dx$ and $v = x$. Therefore,$$\begin{split} V & = \pi \int_0^1 (\ln x)^2 dx \\ & = \pi \left[ x(\ln x)^2 - 2\int x \ln x \frac{1}{x} dx \right]_0^1 \\ & = \pi \left[ x(\ln x)^2 - 2\int \ln x dx \right]_0^1 \\ & = \pi \left[ x(\ln x)^2 - 2(x\ln x - x) \right]_0^1 \\ & = \pi \left[ (\ln 1)^2 - 2(1\ln 1 - 1) - 0 \right] \\ & = \pi \left[ 0 - 2(0 - 1) \right] \\ & = \pi (2) \\ & = 2\pi \end{split}$$Therefore, the volume of the solid is $2\pi$.

Find the Volume of the Solid Obtained by Rotating the Region