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| | <span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves <math style="vertical-align: -4px">y=x</math> and <math style="vertical-align: -4px">y=x^2</math> about the line <math>y=2.</math> | | <span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves <math style="vertical-align: -4px">y=x</math> and <math style="vertical-align: -4px">y=x^2</math> about the line <math>y=2.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009B Sample Final 2, Problem 3 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: 0px">x</math>-axis using the washer method is given by
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| − | <math style="vertical-align: -18px">\int \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx,</math> where <math style="vertical-align: -4px">r_{\text{inner}}</math> is the inner radius of the washer and <math style="vertical-align: -4px">r_{\text{outer}}</math> is the outer radius of the washer.
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| − | '''Solution:''' | + | [[009B Sample Final 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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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 need to find the intersection points of <math style="vertical-align: -5px">y=x</math> and <math style="vertical-align: -5px">y=x^2.</math>
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| − | |To do this, we need to solve
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| − | | <math style="vertical-align: 0px">x=x^2.</math>
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| − | |Moving all the terms on one side of the equation, we get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{0} & = & \displaystyle{x^2-x}\\
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| − | &&\\
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| − | & = & \displaystyle{x(x-1).}
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| − | \end{array}</math>
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| − | |Hence, these two curves intersect at <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: 0px">x=1.</math>
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| − | |So, we are interested in the region between <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: -1px">x=1.</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 use the washer method to calculate this volume.
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| − | |The outer radius is
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| − | | <math style="vertical-align: -4px">r_{\text{outer}}=2-x^2</math>
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| − | |and the inner radius is
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| − | | <math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math>
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| − | |Therefore, the volume of the solid is
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{V} & = & \displaystyle{\int_0^1 \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_0^1 \pi((2-x^2)^2-(2-x)^2)~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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| − | |Now, we integrate to get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{V} & = & \displaystyle{\pi \int_0^1 ((4-4x^2+x^4)-(4-4x+x^2))~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\pi \int_0^1 (4x-5x^2+x^4)~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\pi\bigg(2x^2-\frac{5x^3}{3}+\frac{x^5}{5}\bigg)\bigg|_0^1}\\
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| − | &&\\
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| − | & = & \displaystyle{\pi\bigg(2-\frac{5}{3}+\frac{1}{5}\bigg)-0}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{8\pi}{15}.}
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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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| − | | <math>\frac{8\pi}{15}</math>
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| − | |}
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| | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
