009B Sample Final 1, Problem 7

(a) Find the length of the curve

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a\leq x\leq b}   is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle S=\int 2\pi x\,ds,}   where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle ds={\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.}

Solution:

(a)

Step 1:
First, we calculate  ${\frac {dy}{dx}}.$
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  $y=1-x^{2},$
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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,   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{1}=1+4(0)^{2}=1}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
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{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}}

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)}

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