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

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

is rotated about the 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 y} -axis. Find the area of the resulting surface.

Temp1

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

Solution:

Temp2

(a)

Step 1:
First, we calculate 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{dy}{dx}} .
Since 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 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
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 L=\int_0^{\frac{\pi}{3}} \sqrt{1+(-\tan x)^2}~dx} .

Step 2:
Now, we have:
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} L & = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{1+\tan^2 x}~dx}\\ &&\\ & = & \displaystyle{\int_0^{\frac{\pi}{3}} \sqrt{\sec^2x}~dx}\\ &&\\ & = & \displaystyle{\int_0^{\frac{\pi}{3}} \sec x ~dx}\\ \end{array}}

Step 3:
Finally,
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} 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 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{dy}{dx}} .
Since 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 y=1-x^2,~ \frac{dy}{dx}=-2x} .
Using the formula given in the Foundations section, we have
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 S=\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx} .

Step 2:
Now, we have 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 S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx}
We proceed by using trig substitution. Let 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 x=\frac{1}{2}\tan \theta} . Then, 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 dx=\frac{1}{2}\sec^2\theta d\theta} .
So, we have
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{\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 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 u} -substitution. Let 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 u=\sec \theta} . Then, 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 du=\sec \theta \tan \theta d\theta} .
So, the integral becomes
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{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\ &&\\ & = & \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
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} 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) 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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