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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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| − | |'''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.''' Recall | |
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| − | | <math style="vertical-align: -14px">\int \sec x~dx=\ln|\sec(x)+\tan(x)|+C.</math>
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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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| − | <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}~dx.</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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| − | |First, we calculate <math>\frac{dy}{dx}.</math>
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| − | |Since <math style="vertical-align: -5px">y=\ln (\cos x),</math>
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| − | | <math>\frac{dy}{dx}=\frac{1}{\cos x}(-\sin x)=-\tan x.</math>
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| − | |Using the formula given in the Foundations section, we have
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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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| − | |Now, we have
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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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| − | {| 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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| − | <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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| − | '''(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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| − | |Since <math style="vertical-align: -5px">y=1-x^2,</math>
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| − | | <math>\frac{dy}{dx}=-2x.</math>
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| − | |Using the formula given in the Foundations section, we have
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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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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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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| − | |We proceed by <math>u</math>-substitution.
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| − | |Let <math style="vertical-align: -2px">u=1+4x^2.</math>
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| − | |Then, <math style="vertical-align: 0px">du=8xdx</math> and <math style="vertical-align: -14px">\frac{du}{8}=xdx.</math>
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| − | |-
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| − | |Since the integral is a definite integral, we need to change the bounds of integration.
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| − | |Plugging in our values into the equation <math style="vertical-align: -4px">u=1+4x^2,</math> we get
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| − | | <math style="vertical-align: -5px">u_1=1+4(0)^2=1</math> and <math style="vertical-align: -5px">u_2=1+4(1)^2=5.</math>
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| − | |-
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| − | |Thus, the integral becomes
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| − | <math>\begin{array}{rcl}
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| − | S& = & \displaystyle{\int_1^5 \frac{2\pi}{8} \sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{4} \int_1^5 u^{\frac{1}{2}}~du.}
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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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| − | !Step 3:
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| − | |-
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| − | |Now, we integrate to get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{S} & = & \displaystyle{\frac{\pi}{4}\bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{1}^{5}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}u^{\frac{3}{2}}\bigg|_{1}^{5}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}(5)^{\frac{3}{2}}-\frac{\pi}{6}(1)^{\frac{3}{2}}}\\
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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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| − | | '''(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>''']] |
