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:
The formula for the length of a curve $y=f(x)$ where $a\leq x\leq b$ is $L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx$.

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

(b)

Step 1:

Step 2:

Step 3:

Final Answer:
(a) $\ln(2+{\sqrt {3}})$
(b)

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