Difference between revisions of "009B Sample Final 1, Problem 7"

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::<span class="exam">a) Find the length of the curve
+
<span class="exam">(a) Find the length of the curve
  
−
::::::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}.</math>
+
::<math>y=\ln (\cos x),~~~0\leq x \leq \frac{\pi}{3}</math>.
  
−
::<span class="exam">b) The curve
+
<span class="exam">(b) The curve
  
−
::::::<math>y=1-x^2,~~~0\leq x \leq 1</math>
+
::<math>y=1-x^2,~~~0\leq x \leq 1</math>
  
−
:::<span class="exam">is rotated about the <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
+
<span class="exam">is rotated about the &nbsp;<math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface.
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|Recall:
+
|'''1.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is
 
|-
 
|-
 
|
 
|
−
::'''1.''' The formula for the length <math style="vertical-align: 0px">L</math> of a curve <math style="vertical-align: -5px">y=f(x)</math> where <math style="vertical-align: -3px">a\leq x \leq b</math> is
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>  
 
|-
 
|-
−
|
+
|'''2.''' Recall
−
:::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
−
::'''2.''' <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
 
 
|-
 
|-
−
|
+
|'''3.''' The surface area &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; of a function &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; rotated about the &nbsp;<math style="vertical-align: -4px">y</math>-axis is given by  
−
::'''3.''' The surface area <math style="vertical-align: 0px">S</math> of a function <math style="vertical-align: -5px">y=f(x)</math> rotated about the <math style="vertical-align: -4px">y</math>-axis is given by  
 
 
|-
 
|-
 
|
 
|
−
:::<math style="vertical-align: -13px">S=\int 2\pi x\,ds</math>, where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
Line 37: Line 35:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we calculate&thinsp; <math>\frac{dy}{dx}.</math>  
+
|First, we calculate &nbsp;<math>\frac{dy}{dx}.</math>  
 
|-
 
|-
−
|Since <math style="vertical-align: -5px">y=\ln (\cos x),</math>
+
|Since &nbsp;<math style="vertical-align: -5px">y=\ln (\cos x),</math>
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x.</math>
−
::<math>\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x.</math>
 
 
|-
 
|-
 
|Using the formula given in the Foundations section, we have
 
|Using the formula given in the Foundations section, we have
 
|-
 
|-
 
|
 
|
−
::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we have:
+
|Now, we have
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
 
L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
L& = & \ln |\sec x+\tan x|\bigg|_0^{\frac{\pi}{3}}\\
 
L& = & \ln |\sec x+\tan x|\bigg|_0^{\frac{\pi}{3}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\ln (2+\sqrt{3})}.
 
& = & \displaystyle{\ln (2+\sqrt{3})}.
 
\end{array}</math>
 
\end{array}</math>
 +
|-
 +
|
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|We start by calculating&thinsp; <math>\frac{dy}{dx}.</math>  
+
|We start by calculating &nbsp;<math>\frac{dy}{dx}.</math>  
 
|-
 
|-
−
|Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x.</math>
+
|Since &nbsp;<math style="vertical-align: -5px">y=1-x^2,</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=-2x.</math>
 
|-
 
|-
 
|Using the formula given in the Foundations section, we have
 
|Using the formula given in the Foundations section, we have
 
|-
 
|-
 
|
 
|
−
::<math>S\,=\,\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>S\,=\,\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we have <math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
+
|Now, we have &nbsp;<math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
 
|-
 
|-
−
|We proceed by using trig substitution. Let <math style="vertical-align: -13px">x=\frac{1}{2}\tan \theta.</math> Then, <math style="vertical-align:  -12px">dx=\frac{1}{2}\sec^2\theta \,d\theta.</math>
+
|We proceed by &nbsp;<math>u</math>-substitution.  
 
|-
 
|-
−
|So, we have
+
|Let &nbsp;<math style="vertical-align: -2px">u=1+4x^2.</math> &nbsp;
 
|-
 
|-
−
|
+
|Then, &nbsp; <math style="vertical-align: 0px">du=8xdx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{8}=xdx.</math>
−
::<math>\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}\\
+
|Since the integral is a definite integral, we need to change the bounds of integration.
−
&&\\
+
|-
−
& = & \displaystyle{\int \frac{\pi}{2} \tan \theta \sec \theta \sec^2\theta d\theta}.\\
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=1+4x^2,</math>&nbsp; we get
−
\end{array}</math>
 
−
|}
 
−
 
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 3: &nbsp;
 
 
|-
 
|-
−
|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=\sec \theta.</math> Then, <math style="vertical-align: -1px">du=\sec \theta \tan \theta \,d\theta.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=1+4(0)^2=1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=1+4(1)^2=5.</math>
 
|-
 
|-
−
|So, the integral becomes
+
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
−
\displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\
+
S& = & \displaystyle{\int_1^5 \frac{2\pi}{8} \sqrt{u}~du}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{\pi}{6}u^3+C}\\
+
& = & \displaystyle{\frac{\pi}{4} \int_1^5 u^{\frac{1}{2}}~du.}
−
&&\\
 
−
& = & \displaystyle{\frac{\pi}{6}\sec^3\theta+C}\\
 
−
&&\\
 
−
& = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3+C}.\\
 
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
−
!Step 4: &nbsp;
+
!Step 3: &nbsp;
 
|-
 
|-
−
|We started with a definite integral. So, using Step 2 and 3, we have
+
|Now, we integrate to get
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
−
S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\
+
\displaystyle{S} & = & \displaystyle{\frac{\pi}{4}\bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{1}^{5}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3}\bigg|_0^1\\
+
& = & \displaystyle{\frac{\pi}{6}u^{\frac{3}{2}}\bigg|_{1}^{5}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\frac{\pi(\sqrt{5})^3}{6}-\frac{\pi}{6}}\\
+
& = & \displaystyle{\frac{\pi}{6}(5)^{\frac{3}{2}}-\frac{\pi}{6}(1)^{\frac{3}{2}}}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}.\\
 
& = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; '''(a)''' &nbsp;<math>\ln (2+\sqrt{3})</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>\ln (2+\sqrt{3})</math>
 
|-
 
|-
−
|&nbsp;&nbsp; '''(b)''' &nbsp;<math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:06, 20 May 2017

(a) Find the length of 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=\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.

Foundations:  
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. Recall
        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}~dx.}


Solution:

(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),}
       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}=\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^{\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^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\ &&\\ & = & \displaystyle{\int_0^{\pi/3} \sqrt{\sec^2x}~dx}\\ &&\\ & = & \displaystyle{\int_0^{\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}}

(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,}
       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}=-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  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=1+4x^2.}  
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=8xdx}   and  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{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  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=1+4x^2,}   we 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 u_1=1+4(0)^2=1}   and  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_2=1+4(1)^2=5.}
Thus, 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} 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)}

Return to Sample Exam

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