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.

Temp1

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

Solution:

Temp2

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

Temp3

(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
${\begin{array}{rcl}\displaystyle {\int 2\pi x{\sqrt {1+4x^{2}}}~dx}&=&\displaystyle {\int {\frac {\pi }{2}}u^{2}du}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}u^{3}+C}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}\sec ^{3}\theta +C}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}({\sqrt {1+4x^{2}}})^{3}+C}\\\end{array}}$

Step 4:
We started with a definite integral. So, using Step 2 and 3, we have
${\begin{array}{rcl}S&=&\displaystyle {\int _{0}^{1}2\pi x{\sqrt {1+4x^{2}}}~dx}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}({\sqrt {1+4x^{2}}})^{3}}{\bigg |}_{0}^{1}\\&&\\&=&\displaystyle {{\frac {\pi ({\sqrt {5}})^{3}}{6}}-{\frac {\pi }{6}}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}(5{\sqrt {5}}-1)}\\\end{array}}$

Temp4

Final Answer:
(a) $\ln(2+{\sqrt {3}})$
(b) ${\frac {\pi }{6}}(5{\sqrt {5}}-1)$

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