009B Sample Final 1, Problem 5 - Revision history

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MathAdmin: Created page with " Consider the solid obtained by rotating the area bounded by the following three functions about the $y$ -axis: ::::..."

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Created page with " Consider the solid obtained by rotating the area bounded by the following three functions about the <math style="vertical-align: -3px">y</math>-axis: ::::..."

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Consider the solid obtained by rotating the area bounded by the following three functions about the <math style="vertical-align: -3px">y</math>-axis:

:::::: <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>.

a) Sketch the region bounded by the given three functions. Find the intersection point of the two functions:

:<math style="vertical-align: -4px">y=e^x</math> and <math style="vertical-align: -4px">y=ex</math>. (There is only one.)

b) Set up the integral for the volume of the solid.

c) Find the volume of the solid by computing the integral.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math>
|-
|
::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x</math>.
|-
|'''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
|-
|
::<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.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we sketch the region bounded by the three functions.
|-
|Insert graph here.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Setting the equations equal, we have <math style="vertical-align: 0px">e^x=ex</math>.
|-
|We get one intersection point, which is <math style="vertical-align: -4px">(1,e)</math>.
|-
|This intersection point can be seen in the graph shown in Step 1.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We proceed using cylindrical shells. The radius of the shells is given by <math style="vertical-align: 0px">r=x</math>.
|-
|The height of the shells is given by <math style="vertical-align: 0px">h=e^x-ex</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the volume of the solid is
|-
|
::<math style="vertical-align: -14px">\int 2\pi rh\,dx\,=\,\int_0^1 2\pi x(e^x-ex)\,dx.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We need to integrate
|-
|
::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|For the first integral, we need to use integration by parts.
|-
|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>.
|-
|So, the integral becomes
|-
|
::<math>\begin{array}{rcl}
\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}\\
&&\\
& = & \displaystyle{2\pi\bigg(xe^x-e^x\bigg)\bigg|_0^1-\frac{2\pi e}{3}}\\
&&\\
& = & \displaystyle{2\pi(e-e-(-1))-\frac{2\pi e}{3}}\\
&&\\
& = & \displaystyle{2\pi-\frac{2\pi e}{3}}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -5px">(1,e)</math> (See Step 1 for the graph)
|-
|'''(b)'''  <math style="vertical-align: -15px">\int_0^1 2\pi x(e^x-ex)~dx</math>
|-
|'''(c)'''  <math style="vertical-align: -14px">2\pi-\frac{2\pi e}{3}</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 31: / Line 31: @@
 |First, we sketch the region bounded by the given functions.
 |-
-|[[File:009A_SF1_5.png|center|300px]]
+|[[File:009B_SF1_5.png|center|300px]]
 |}

@@ Line 31: / Line 31: @@
 |First, we sketch the region bounded by the given functions.
 |-
-|Insert graph here.
+|[[File:009A_SF1_5.png|center|300px]]
 |}

@@ Line 35: / Line 35: @@
 |-
 |First, we sketch the region bounded by the three functions.  The region is shown in red, while the revolved solid is shown in blue.
 |-
 |[[File:9BF1 5 GP.png|center|500px]]

@@ Line 110: / Line 110: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -5px">(1,e)</math> (See Step 1 for the graph)
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -5px">(1,e)</math> (See Step 1 for the graph)
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -15px">\int_0^1 2\pi x(e^x-ex)~dx</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: -15px">\int_0^1 2\pi x(e^x-ex)~dx</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: -14px">2\pi-\frac{2\pi e}{3}</math>
+|&nbsp;&nbsp; '''(c)''' &nbsp;<math style="vertical-align: -14px">2\pi-\frac{2\pi e}{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::::::<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 three functions. Find the intersection point of the two functions:
+::<span class="exam">a) Sketch the region bounded by the given three functions. Find the intersection point of the two functions:
-:<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"><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">b) Set up the integral for the volume of the solid.
+::<span class="exam">b) Set up the integral for the volume of the solid.
-<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.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 16: / Line 16: @@
 |Recall:
 |-
 |'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math>
 |-
+|
-::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x</math>.
+:::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x.</math>
 |-
 |'''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
 |-
+|
-::<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.
+:::<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.
 |}
@@ Line 44: / Line 46: @@
 !Step 2: &nbsp;
 |-
-|Setting the equations equal, we have <math style="vertical-align: 0px">e^x=ex</math>.
+|Setting the equations equal, we have <math style="vertical-align: 0px">e^x=ex.</math>
 |-
-|We get one intersection point, which is <math style="vertical-align: -4px">(1,e)</math>.
+|We get one intersection point, which is <math style="vertical-align: -4px">(1,e).</math>
 |-
 |This intersection point can be seen in the graph shown in Step 1.
@@ Line 56: / Line 58: @@
 !Step 1: &nbsp;
 |-
-|We proceed using cylindrical shells. The radius of the shells is given by <math style="vertical-align: 0px">r=x</math>.
+|We proceed using cylindrical shells. The radius of the shells is given by <math style="vertical-align: 0px">r=x.</math>
 |-
-|The height of the shells is given by <math style="vertical-align: 0px">h=e^x-ex</math>.
 |}
@@ Line 86: / Line 91: @@
 |For the first integral, we need to use integration by parts.
 |-
-|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>.
+|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>
 |-
 |So, the integral becomes