# 009B Sample Final 1, Problem 7

(a) Find the length of the curve

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

(b) The curve

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

is rotated about the  ${\displaystyle y}$-axis. Find the area of the resulting surface.

Foundations:
1. The formula for the length  ${\displaystyle L}$  of a curve  ${\displaystyle y=f(x)}$  where  ${\displaystyle a\leq x\leq b}$  is

${\displaystyle L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.}$

2. Recall
${\displaystyle \int \sec x~dx=\ln |\sec(x)+\tan(x)|+C.}$
3. The surface area  ${\displaystyle S}$  of a function  ${\displaystyle y=f(x)}$  rotated about the  ${\displaystyle y}$-axis is given by

${\displaystyle S=\int 2\pi x\,ds,}$  where ${\displaystyle ds={\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.}$

Solution:

(a)

Step 1:
First, we calculate  ${\displaystyle {\frac {dy}{dx}}.}$
Since  ${\displaystyle y=\ln(\cos x),}$
${\displaystyle {\frac {dy}{dx}}={\frac {1}{\cos x}}(-\sin x)=-\tan x.}$
Using the formula given in the Foundations section, we have

${\displaystyle L=\int _{0}^{\pi /3}{\sqrt {1+(-\tan x)^{2}}}~dx.}$

Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}L&=&\displaystyle {\int _{0}^{\pi /3}{\sqrt {1+\tan ^{2}x}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi /3}{\sqrt {\sec ^{2}x}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi /3}\sec x~dx}.\\\end{array}}}$

Step 3:
Finally,

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

(b)

Step 1:
We start by calculating  ${\displaystyle {\frac {dy}{dx}}.}$
Since  ${\displaystyle y=1-x^{2},}$
${\displaystyle {\frac {dy}{dx}}=-2x.}$
Using the formula given in the Foundations section, we have

${\displaystyle S\,=\,\int _{0}^{1}2\pi x{\sqrt {1+(-2x)^{2}}}~dx.}$

Step 2:
Now, we have  ${\displaystyle S=\int _{0}^{1}2\pi x{\sqrt {1+4x^{2}}}~dx.}$
We proceed by  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=1+4x^{2}.}$
Then,   ${\displaystyle du=8xdx}$  and  ${\displaystyle {\frac {du}{8}}=xdx.}$
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation  ${\displaystyle u=1+4x^{2},}$  we get
${\displaystyle u_{1}=1+4(0)^{2}=1}$  and  ${\displaystyle u_{2}=1+4(1)^{2}=5.}$
Thus, the integral becomes

${\displaystyle {\begin{array}{rcl}S&=&\displaystyle {\int _{1}^{5}{\frac {2\pi }{8}}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{4}}\int _{1}^{5}u^{\frac {1}{2}}~du.}\end{array}}}$

Step 3:
Now, we integrate to get

${\displaystyle {\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {{\frac {\pi }{4}}{\bigg (}{\frac {2}{3}}u^{\frac {3}{2}}{\bigg )}{\bigg |}_{1}^{5}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}u^{\frac {3}{2}}{\bigg |}_{1}^{5}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}(5)^{\frac {3}{2}}-{\frac {\pi }{6}}(1)^{\frac {3}{2}}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}(5{\sqrt {5}}-1)}.\\\end{array}}}$

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