009B Sample Final 1, Problem 7

a) Find the length of the curve

$y=\ln(\cos x),~~~0\leq x\leq {\frac {\pi }{3}}$.

b) The curve

$y=1-x^{2},~~~0\leq x\leq 1$

is rotated about the $y$-axis. Find the area of the resulting surface.

Foundations:
1. The formula for the length $L$ of a curve $y=f(x)$ where $a\leq x\leq b$ is
$L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx$.
2. Recall that $\int \sec x~dx=\ln |\sec(x)+\tan(x)|+C$.
3. The surface area $S$ of a function $y=f(x)$ rotated about the $y$-axis is given by
$S=\int 2\pi xds$ where $ds={\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}$.

Solution:

(a)

Step 1:
First, we calculate ${\frac {dy}{dx}}$.
Since $y=\ln(\cos x),~{\frac {dy}{dx}}={\frac {1}{\cos x}}(-\sin x)=-\tan x$.
Using the formula given in the Foundations section, we have
$L=\int _{0}^{\frac {\pi }{3}}{\sqrt {1+(-\tan x)^{2}}}~dx$.

Step 2:
Now, we have:
${\begin{array}{rcl}L&=&\displaystyle {\int _{0}^{\frac {\pi }{3}}{\sqrt {1+\tan ^{2}x}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\frac {\pi }{3}}{\sqrt {\sec ^{2}x}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\frac {\pi }{3}}\sec x~dx}\\\end{array}}$

Step 3:
Finally,
${\begin{array}{rcl}L&=&\ln |\sec x+\tan x|{\bigg |}_{0}^{\frac {\pi }{3}}\\&&\\&=&\displaystyle {\ln {\bigg |}\sec {\frac {\pi }{3}}+\tan {\frac {\pi }{3}}{\bigg |}-\ln |\sec 0+\tan 0|}\\&&\\&=&\displaystyle {\ln |2+{\sqrt {3}}|-\ln |1|}\\&&\\&=&\displaystyle {\ln(2+{\sqrt {3}})}\end{array}}$

(b)

Step 1:
We start by calculating ${\frac {dy}{dx}}$.
Since $y=1-x^{2},~{\frac {dy}{dx}}=-2x$.
Using the formula given in the Foundations section, we have
$S=\int _{0}^{1}2\pi x{\sqrt {1+(-2x)^{2}}}~dx$.

Step 2:
Now, we have $S=\int _{0}^{1}2\pi x{\sqrt {1+4x^{2}}}~dx$
We proceed by using trig substitution. Let $x={\frac {1}{2}}\tan \theta$. Then, $dx={\frac {1}{2}}\sec ^{2}\theta d\theta$.
So, we have
${\begin{array}{rcl}\displaystyle {\int 2\pi x{\sqrt {1+4x^{2}}}~dx}&=&\displaystyle {\int 2\pi {\bigg (}{\frac {1}{2}}\tan \theta {\bigg )}{\sqrt {1+\tan ^{2}\theta }}{\bigg (}{\frac {1}{2}}\sec ^{2}\theta {\bigg )}d\theta }\\&&\\&=&\displaystyle {\int {\frac {\pi }{2}}\tan \theta \sec \theta \sec ^{2}\theta d\theta }\\\end{array}}$

Step 3:
Now, we use $u$-substitution. Let $u=\sec \theta$. Then, $du=\sec \theta \tan \theta d\theta$.
So, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}u^3+C}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}\sec^3\theta+C}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3+C}\\ \end{array}}

Step 4:
We started with a definite integral. So, using Step 2 and 3, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3}\bigg|_0^1\\ &&\\ & = & \displaystyle{\frac{\pi(\sqrt{5})^3}{6}-\frac{\pi}{6}}\\ &&\\ & = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}\\ \end{array}}

Final Answer:
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln (2+\sqrt{3})}
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pi}{6}(5\sqrt{5}-1)}

Return to Sample Exam