009B Sample Final 1, Problem 5 - Revision history

Kayla Murray at 00:17, 3 December 2017

2017-12-03T00:17:30Z

Kayla Murray at 19:45, 18 March 2017

2017-03-18T19:45:08Z

Kayla Murray at 15:54, 28 February 2017

2017-02-28T15:54:44Z

Kayla Murray at 15:48, 28 February 2017

2017-02-28T15:48:48Z

Kayla Murray at 15:37, 28 February 2017

2017-02-28T15:37:56Z

Kayla Murray at 15:30, 28 February 2017

2017-02-28T15:30:59Z

Kayla Murray at 20:13, 27 February 2017

2017-02-27T20:13:12Z

Kayla Murray at 02:12, 19 February 2017

2017-02-19T02:12:14Z

Grad at 07:07, 26 February 2016

2016-02-26T07:07:41Z

Grad: /* 5 */

2016-02-26T07:07:32Z

@@ Line 7: / Line 7: @@
 <span class="exam">(c) Find the volume of the solid by computing the integral.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 53: / Line 53: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,4).</math>
 |-
-|This intersection point can be seen in the graph shown in Step 1.
+|This intersection points can be seen in the graph shown in Step 1.
 |}
@@ Line 107: / Line 107: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</math> (See Step 1 for the graph)
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</math>&nbsp; (See Step 1 for the graph)
 |-
 |&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>

@@ Line 39: / Line 39: @@
 |Setting the equations equal, we have &nbsp;<math style="vertical-align: 0px">x^2=2x.</math>
 |-
-|Solving for &nbsp;<math>x,</math>&nbsp; we get
+|Solving for &nbsp;<math style="vertical-align: -4px">x,</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
@@ Line 47: / Line 47: @@
 \end{array}</math>
 |-
-|So, &nbsp;<math>x=0</math>&nbsp; and &nbsp;<math>x=2.</math>
+|So, &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2.</math>
 |-
 |If we plug these values into our functions, we get the intersection points
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>(0,0)</math>&nbsp; and &nbsp;<math>(2,4).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,4).</math>
 |-
 |This intersection point can be seen in the graph shown in Step 1.
@@ Line 107: / Line 107: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</math> (See Step 1 for the graph)
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</math> (See Step 1 for the graph)
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: -14px">\frac{8\pi}{3}</math>
+|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;<math style="vertical-align: -14px">\frac{8\pi}{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 37: / Line 37: @@
 !Step 2: &nbsp;
 |-
-|Setting the equations equal, we have &nbsp;<math style="vertical-align: 0px">x^2=2x</math>.
+|Setting the equations equal, we have &nbsp;<math style="vertical-align: 0px">x^2=2x.</math>
 |-
 |Solving for &nbsp;<math>x,</math>&nbsp; we get
@@ Line 61: / Line 61: @@
 !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 &nbsp;<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>.
+|The height of the shells is given by &nbsp;<math style="vertical-align: 0px">h=2x-x^2.</math>
 |}
@@ Line 72: / Line 72: @@
 |-
+|
-::<math style="vertical-align: -14px">\int 2\pi rh\,dx\,=\,\int_0^1 2\pi x(e^x-ex)\,dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\int 2\pi rh~dx~=~\int_0^2 2\pi x(2x-x^2)~dx.</math>
 |}
@@ Line 83: / Line 83: @@
 |-
+|
-::<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>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^2 2\pi x(2x-x^2)~dx~=~2\pi\int_0^2 2x^2-x^3~dx.</math>
 |}
@@ Line 89: / Line 89: @@
 !Step 2: &nbsp;
 |-
-|For the first integral, we need to use integration by parts.
+|We have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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{\int_0^2 2\pi x(2x-x^2)~dx} & = & \displaystyle{2\pi\int_0^2 2x^2-x^3~dx}\\
 &&\\
-& = & \displaystyle{2\pi\bigg(xe^x-e^x\bigg)\bigg|_0^1-\frac{2\pi e}{3}}\\
+& = & \displaystyle{2\pi\bigg(\frac{2x^3}{3}-\frac{x^4}{4}\bigg)\bigg|_0^2}\\
 &&\\
-& = & \displaystyle{2\pi(e-e-(-1))-\frac{2\pi e}{3}}\\
+& = & \displaystyle{2\pi\bigg(\frac{2^4}{3}-\frac{2^4}{4}\bigg)-2\pi(0)}\\
 &&\\
-& = & \displaystyle{2\pi-\frac{2\pi e}{3}}.\\
+& = & \displaystyle{\frac{8\pi}{3}.}\\
 \end{array}</math>
 |}
@@ Line 113: / Line 109: @@
 |'''(a)''' &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</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>
+|'''(b)''' &nbsp;<math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: -14px">2\pi-\frac{2\pi e}{3}</math>
+|'''(c)''' &nbsp;<math style="vertical-align: -14px">\frac{8\pi}{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 29: / Line 29: @@
 !Step 1: &nbsp;
 |-
-|First, we sketch the region bounded by the three functions.
+|First, we sketch the region bounded by the given functions.
 |-
 |Insert graph here.
@@ Line 37: / Line 37: @@
 !Step 2: &nbsp;
 |-
-|Setting the equations equal, we have <math style="vertical-align: 0px">e^x=ex</math>.
+|Setting the equations equal, we have &nbsp;<math style="vertical-align: 0px">x^2=2x</math>.
 |-
-|We get one intersection point, which is <math style="vertical-align: -4px">(1,e)</math>.
+|Solving for &nbsp;<math>x,</math>&nbsp; we get
 |-
 |This intersection point can be seen in the graph shown in Step 1.
@@ Line 99: / Line 111: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -5px">(1,e)</math> (See Step 1 for the graph)
+|'''(a)''' &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</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>

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.''' You can find the intersection points of two functions, say &nbsp; <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>.
+&nbsp; &nbsp; &nbsp; &nbsp; by setting &nbsp;<math style="vertical-align: -5px">f(x)=g(x)</math>&nbsp; and solving for &nbsp;<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
+|'''2.''' The volume of a solid obtained by rotating an area around the &nbsp;<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.
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int 2\pi rh~dx,</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; is the radius of the shells and &nbsp;<math style="vertical-align: 0px">h</math>&nbsp; is the height of the shells.
 |}

@@ Line 1: / Line 1: @@
-<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 &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; and the line &nbsp;<math style="vertical-align: -4px">y=2x</math>&nbsp; in the first quadrant is revolved about the &nbsp;<math style="vertical-align: -4px">y</math>-axis to generate a solid.
-::<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.
 <span class="exam">(b) Set up the integral for the volume of the solid.