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| | ::<math>y=1-x^2,~~~0\leq x \leq 1</math> | | ::<math>y=1-x^2,~~~0\leq x \leq 1</math> |
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| − | <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 <math style="vertical-align: -3px">y</math>-axis. Find the area of the resulting surface. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']] |
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| − | |Recall:
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| − | |'''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
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| − | ::<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
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| − | |- | |
| − | |'''2.''' <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
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| − | |-
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| − | |'''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
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| − | |-
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| − | ::<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>
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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we calculate  <math>\frac{dy}{dx}.</math>
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| − | |Since <math style="vertical-align: -12px">y=\ln (\cos x),~\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x</math>.
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| − | |-
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| − | |Using the formula given in the Foundations section, we have
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| − | |-
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| − | |
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| − | ::<math>L=\int_0^{\pi/3} \sqrt{1+(-\tan x)^2}~dx</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we have:
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | L & = & \displaystyle{\int_0^{\pi/3} \sqrt{1+\tan^2 x}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_0^{\pi/3} \sqrt{\sec^2x}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_0^{\pi/3} \sec x ~dx}.\\
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| − | \end{array}</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Finally,
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | L& = & \ln |\sec x+\tan x|\bigg|_0^{\frac{\pi}{3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln \bigg|\sec \frac{\pi}{3}+\tan \frac{\pi}{3}\bigg|-\ln|\sec 0 +\tan 0|}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln |2+\sqrt{3}|-\ln|1|}\\
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| − | &&\\
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| − | & = & \displaystyle{\ln (2+\sqrt{3})}.
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| − | \end{array}</math>
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| − | |-
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| − | |
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We start by calculating  <math>\frac{dy}{dx}</math> .
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| − | |-
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| − | |Since <math style="vertical-align: -13px">y=1-x^2,~ \frac{dy}{dx}=-2x</math>.
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| − | |-
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| − | |Using the formula given in the Foundations section, we have
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| − | |-
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| − | |
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| − | ::<math>S\,=\,\int_0^{1}2\pi x \sqrt{1+(-2x)^2}~dx.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we have <math style="vertical-align: -14px">S=\int_0^{1}2\pi x \sqrt{1+4x^2}~dx.</math>
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| − | |-
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| − | |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>.
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| − | |-
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| − | |So, we have
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \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}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \frac{\pi}{2} \tan \theta \sec \theta \sec^2\theta d\theta}.\\
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |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>.
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| − | |So, the integral becomes
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int 2\pi x \sqrt{1+4x^2}~dx} & = & \displaystyle{\int \frac{\pi}{2}u^2du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}u^3+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}\sec^3\theta+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3+C}.\\
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |-
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| − | |We started with a definite integral. So, using Step 2 and 3, we have
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | S & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+4x^2}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}(\sqrt{1+4x^2})^3}\bigg|_0^1\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi(\sqrt{5})^3}{6}-\frac{\pi}{6}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}(5\sqrt{5}-1)}.\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | |'''(a)''' <math>\ln (2+\sqrt{3})</math>
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| − | |-
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| − | |'''(b)''' <math>\frac{\pi}{6}(5\sqrt{5}-1)</math>
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| − | |}
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| | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
