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| − | <span class="exam"> Find the volume of the solid obtained by rotating about the <math>x</math>-axis the region bounded by <math style="vertical-align: -4px">y=\sqrt{1-x^2}</math> and <math>y=0.</math> | + | <span class="exam"> Find the volume of the solid obtained by rotating about the <math>x</math>-axis the region bounded by <math style="vertical-align: -4px">y=\sqrt{1-x^2}</math> and <math style="vertical-align: -4px">y=0.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009B Sample Final 3, Problem 4 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 a region around the <math style="vertical-align: 0px">x</math>-axis using disk method is given by
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| − | <math style="vertical-align: -13px">\int \pi r^2~dx,</math> where <math style="vertical-align: 0px">r</math> is the radius of the disk.
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| − | '''Solution:''' | + | [[009B Sample Final 3, Problem 4 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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| − | |We start by finding the intersection points of the functions <math>y=\sqrt{1-x^2}</math> and <math>y=0.</math>
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| − | |We need to solve <math>0=\sqrt{1-x^2}.</math>
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| − | |If we square both sides, we get
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| − | |<math>0=1-x^2.</math>
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| − | |The solutions to this equation are <math>x=-1</math> and <math>x=1.</math>
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| − | |Hence, we are interested in the region between <math>x=-1</math> and <math>x=1.</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Using the disk method, the radius of each disk is given by <math>r=\sqrt{1-x^2}.</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_{-1}^1 \pi (\sqrt{1-x^2})^2~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_{-1}^1 \pi (1-x^2)~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\pi\bigg(x-\frac{x^3}{3}\bigg)\bigg|_{-1}^1}\\
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| − | &&\\
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| − | & = & \displaystyle{\pi\bigg(1-\frac{1}{3}\bigg)-\pi\bigg(-1+\frac{1}{3}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4\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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| − | | <math>\frac{4\pi}{3}</math>
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| − | |}
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| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
