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| | <span class="exam">(c) Find the volume of the solid by computing the integral. | | <span class="exam">(c) Find the volume of the solid by computing the integral. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009B Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math> | |
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| − | by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x.</math>
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| − | |'''2.''' The volume of a solid obtained by rotating an area around the <math style="vertical-align: -4px">y</math>-axis using cylindrical shells is given by
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| − | <math style="vertical-align: -13px">\int 2\pi rh~dx,</math> where <math style="vertical-align: 0px">r</math> is the radius of the shells and <math style="vertical-align: 0px">h</math> is the height of the shells.
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| − | '''Solution:''' | + | [[009B Sample Final 1, Problem 5 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 sketch the region bounded by the given functions.
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| − | |Insert graph here.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Setting the equations equal, we have <math style="vertical-align: 0px">x^2=2x.</math>
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| − | |Solving for <math style="vertical-align: -4px">x,</math> we get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{0} & = & \displaystyle{x^2-2x}\\
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| − | &&\\
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| − | & = & \displaystyle{x(x-2).}
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| − | \end{array}</math>
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| − | |So, <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: 0px">x=2.</math>
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| − | |If we plug these values into our functions, we get the intersection points
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| − | | <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,4).</math>
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| − | |This intersection points can be seen in the graph shown in Step 1.
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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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| − | |We proceed using cylindrical shells. The radius of the shells is given by <math style="vertical-align: 0px">r=x.</math>
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| − | |The height of the shells is given by <math style="vertical-align: 0px">h=2x-x^2.</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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| − | |So, the volume of the solid is
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| − | <math style="vertical-align: -14px">\int 2\pi rh~dx~=~\int_0^2 2\pi x(2x-x^2)~dx.</math>
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| − | |}
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We need to integrate
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| − | <math>\int_0^2 2\pi x(2x-x^2)~dx~=~2\pi\int_0^2 2x^2-x^3~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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| − | |We have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^2 2\pi x(2x-x^2)~dx} & = & \displaystyle{2\pi\int_0^2 2x^2-x^3~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi\bigg(\frac{2x^3}{3}-\frac{x^4}{4}\bigg)\bigg|_0^2}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi\bigg(\frac{2^4}{3}-\frac{2^4}{4}\bigg)-2\pi(0)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{8\pi}{3}.}\\
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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 style="vertical-align: -5px">(0,0),(2,4)</math> (See Step 1 for the graph)
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| − | | '''(b)''' <math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>
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| − | | '''(c)''' <math style="vertical-align: -14px">\frac{8\pi}{3}</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>''']] |
