# 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
$\int \sec x~dx=\ln |\sec(x)+\tan(x)|+C.$ 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 $ds={\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.$ 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

$L=\int _{0}^{\pi /3}{\sqrt {1+(-\tan x)^{2}}}~dx.$ Step 2:
Now, we have

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

${\begin{array}{rcl}L&=&\ln |\sec x+\tan x|{\bigg |}_{0}^{\frac {\pi }{3}}\\&&\\&=&\displaystyle {\ln {\bigg |}\sec {\frac {\pi }{3}}+\tan {\frac {\pi }{3}}{\bigg |}-\ln |\sec 0+\tan 0|}\\&&\\&=&\displaystyle {\ln |2+{\sqrt {3}}|-\ln |1|}\\&&\\&=&\displaystyle {\ln(2+{\sqrt {3}})}.\end{array}}$ (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  $u$ -substitution.
Let  $u=1+4x^{2}.$ Then,   $du=8xdx$ and  ${\frac {du}{8}}=xdx.$ Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation  $u=1+4x^{2},$ we get
$u_{1}=1+4(0)^{2}=1$ and  $u_{2}=1+4(1)^{2}=5.$ Thus, the integral becomes

${\begin{array}{rcl}S&=&\displaystyle {\int _{1}^{5}{\frac {2\pi }{8}}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{4}}\int _{1}^{5}u^{\frac {1}{2}}~du.}\end{array}}$ Step 3:
Now, we integrate to get

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