Solve the problem.
-A region in the first quadrant is bounded above by the curve $y = \tanh x$, below by the $x$-axis, on the left by the $y$-axis, and on the right by the line $x = \ln 5$. Find the volume of the solid generated by revolving the region about the $x$-axis.
A) $2 \pi$
B) $- \frac { 12 } { 13 }$
C) $\pi \left( \ln 5 - \frac { 12 } { 13 } \right)$
D) 0

Question

Quizplus · Accepted Answer

The curve $y = \tanh x$ approaches the $x$-axis and $y$-axis as $x\to -\infty$ and $y\to 0^+$. When $x \ge \ln 5$, it is tangent to the vertical line $x=\ln 5$. Since the curve is symmetric with respect to the $y$-axis, we can focus on the region in the first octant, as shown below.

[asy]
size(180);
import contour;
import graph;
import solids;

real ymin = 0;
real ymax = 0.9999;

add(contour(ymin^2/(2*ymin^2-1) <= x^2-y^2 && x <= log(5), (0.1,0.1), (log(5),ymax), 99), rgb(0.8,0.8,0.8));
draw((0,ymin)--(0,ymax),black+linewidth(1.25),EndArrow(5));
draw((0,0)--(log(5),0),black+linewidth(1.25),EndArrow(5));
draw(arc((0,0),ymax,0,90),black+linewidth(1.25));
draw((0,0)--(log(5),tanh(log(5))),red+linewidth(1.25));
draw((0,ymin)--(log(5),ymin),dashed);

label("$x$",(log(5),0),(2,0));
label("$y$",(0,ymax),(0,2));
label("$0$",(0.1,0.1),NE);
label("$\ln 5$",(log(5),0),S);
label("$y=\tanh x$",(log(5),tanh(log(5))),E);
label("$y=\frac { 1 } { \sqrt { 2 } }$",(0.2,0.2),W);
label("$y=-\frac { 1 } { \sqrt { 2 } }$",(0.2,-0.2),W);
[/asy]

The region is bounded by the curves
$$x^2-y^2 = y\tanh x \quad \text{and} \quad x = \ln 5.$$ Solving for $y$ in $x^2-y^2=y\tanh x$, we get
$$y = -\frac12\tanh x \pm \frac12\sqrt{\tanh^2x+4x^2}.$$ Since $0\le y\le \tanh x$, we take the positive sign and restrict attention to the region lying below the red line $y=-\frac12\tanh x+\frac12\sqrt{\tanh^2x+4x^2}$.

Rotating the region about the $x$-axis generates a solid with cross-sectional area $\pi y^2$ at height $y$. Thus, the desired volume is
\begin{align*}
\int_0^{\ln 5} \pi\left(-\frac 12\tanh x+\frac 12\sqrt{\tanh^2x+4x^2}\right)^2\,dx &= \int_0^{\ln 5} \pi\left(\frac 14-\frac 12\tanh x\sqrt{\tanh^2x+4x^2}+\frac 14(\tanh^2x+4x^2)\right)\,dx \
&= \pi\int_0^{\ln 5} \left(\frac 54x^2-\frac 12\tanh x\sqrt{\tanh^2x+4x^2}+\frac 14\right)\,dx \
&= \pi\left[\frac 35 x^3+\frac 14(\tanh x)\sqrt{\tanh^2x+4x^2}-\frac 12\ln\left|\frac {2(2x-\sqrt{\tanh^2x+4x^2})}{\tanh x+\sqrt{\tanh^2x+4x^2}}\right|\right]_0^{\ln 5} \
&= \boxed{\pi\left(\ln 5-\frac{12}{13}\right)}.
\end{align*}
The last step above follows from the substitution $u = e^{2x}$.
Hence the answer is $\boxed{\textbf{(C)}}$.

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