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| − | <span class="exam"> Consider the solid obtained by rotating the area bounded by the following three functions about the <math style="vertical-align: -3px">y</math>-axis: | + | <span class="exam"> The region bounded by the parabola <math style="vertical-align: -4px">y=x^2</math> and the line <math style="vertical-align: -4px">y=2x</math> in the first quadrant is revolved about the <math style="vertical-align: -4px">y</math>-axis to generate a solid. |
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| − | ::::::<span class="exam"> <math style="vertical-align: 0px">x=0</math>, <math style="vertical-align: -4px">y=e^x</math>, and <math style="vertical-align: -4px">y=ex</math>.
| + | <span class="exam">(a) Sketch the region bounded by the given functions and find their points of intersection. |
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| − | <span class="exam">a) Sketch the region bounded by the given three functions. Find the intersection point of the two functions: | + | <span class="exam">(b) Set up the integral for the volume of the solid. |
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| − | :<span class="exam"><math style="vertical-align: -4px">y=e^x</math> and <math style="vertical-align: -4px">y=ex</math>. (There is only one.)
| + | <span class="exam">(c) Find the volume of the solid by computing the integral. |
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| − | <span class="exam">b) Set up the integral for the volume of the solid. | + | <hr> |
| | + | [[009B Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | <span class="exam">c) Find the volume of the solid by computing the integral.
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| − | == 1 ==
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |Recall:
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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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| − | |-
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| − | |
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| − | ::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solve for <math style="vertical-align: 0px">x</math>.
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| − | |-
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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>\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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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | == 2 ==
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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 sketch the region bounded by the three functions.
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| − | |Insert graph here.
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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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| − | |Setting the equations equal, we have <math style="vertical-align: 0px">e^x=ex</math>.
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| − | |We get one intersection point, which is <math style="vertical-align: -4px">(1,e)</math>.
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| − | |This intersection point can be seen in the graph shown in Step 1.
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| − | |}
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| − | == 3 ==
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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: -1px">h=e^x-ex</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>\int_0^1 2\pi x(e^x-ex)~dx</math>.
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| − | |}
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| − | == 4 ==
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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_0^1 2\pi x(e^x-ex)~dx=2\pi\int_0^1 xe^x~dx-2\pi\int_0^1ex^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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| − | |For the first integral, we need to use integration by parts.
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| − | |-
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| − | |Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x</math>.
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| − | |So, the integral becomes
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^1 2\pi x(e^x-ex)~dx} & = & \displaystyle{2\pi\bigg(xe^x\bigg|_0^1 -\int_0^1 e^xdx\bigg)-\frac{2\pi ex^3}{3}\bigg|_0^1}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi\bigg(xe^x-e^x\bigg)\bigg|_0^1-\frac{2\pi e}{3}}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi(e-e-(-1))-\frac{2\pi e}{3}}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi-\frac{2\pi e}{3}}\\
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| − | \end{array}</math>
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| − | |}
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| − | == 5 ==
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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">(1,e)</math> (See Step 1 for the graph)
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| − | |'''(b)''' <math style="vertical-align: -5px">\int_0^1 2\pi x(e^x-ex)~dx</math>
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| − | |'''(c)''' <math style="vertical-align: -14px">2\pi-\frac{2\pi e}{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>''']] |
