009B Sample Final 2, Problem 3 - Revision history

Kayla Murray at 00:26, 3 December 2017

2017-12-03T00:26:39Z

Kayla Murray at 21:29, 12 March 2017

2017-03-12T21:29:44Z

Kayla Murray at 23:21, 4 March 2017

2017-03-04T23:21:37Z

Kayla Murray at 23:13, 4 March 2017

2017-03-04T23:13:39Z

Kayla Murray at 20:47, 3 March 2017

2017-03-03T20:47:06Z

Kayla Murray at 02:22, 19 February 2017

2017-02-19T02:22:05Z

Kayla Murray at 03:10, 18 February 2017

2017-02-18T03:10:47Z

Kayla Murray: Created page with "Consider the region bounded by the following two functions: :::::::: $y=2(-x^2+9)$ and
2017-02-18T02:43:30Z

Created page with "Consider the region bounded by the following two functions: :::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math styl..."

New page
Consider the region bounded by the following two functions:
:::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math style="vertical-align: -4px">y=0</math>.

a) Using the lower sum with three rectangles having equal width, approximate the area.

b) Using the upper sum with three rectangles having equal width, approximate the area.

c) Find the actual area of the region.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|
|-
|
|-
|
|-
|
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|-
|
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''
|-
|'''(b)'''
|-
|'''(c)'''
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves &nbsp;<math style="vertical-align: -4px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; about the line &nbsp;<math>y=2.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 2, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 23: / Line 23: @@
 |First, we need to find the intersection points of &nbsp;<math style="vertical-align: -5px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=x^2.</math>
 |-
-|To do this, we need to solve &nbsp;<math style="vertical-align: 0px">x=x^2.</math>
+|To do this, we need to solve
 |-
 |Moving all the terms on one side of the equation, we get
@@ Line 43: / Line 45: @@
 |We use the washer method to calculate this volume.
 |-
-|The outer radius is &nbsp;<math style="vertical-align: -4px">r_{\text{outer}}=2-x^2</math>&nbsp; and
+|The outer radius is
 |-
-|the inner radius is &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math>&nbsp;
 |-
 |Therefore, the volume of the solid is

@@ Line 9: / Line 9: @@
 &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 &nbsp;<math style="vertical-align: -4px">x</math>-axis using the washer method is given by
+|'''2.''' The volume of a solid obtained by rotating an area around the &nbsp;<math style="vertical-align: 0px">x</math>-axis using the washer method is given by
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx,</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">r_{\text{inner}}</math>&nbsp; is the inner radius of the washer and &nbsp;<math style="vertical-align: 0px">r_{\text{outer}}</math>&nbsp; is the outer radius of the washer.
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">\int \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx,</math>&nbsp; where &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}</math>&nbsp; is the inner radius of the washer and &nbsp;<math style="vertical-align: -4px">r_{\text{outer}}</math>&nbsp; is the outer radius of the washer.
 |}
@@ Line 21: / Line 21: @@
 !Step 1: &nbsp;
 |-
-|First, we need to find the intersection points of &nbsp;<math>y=x</math>&nbsp; and &nbsp;<math>y=x^2.</math>
+|First, we need to find the intersection points of &nbsp;<math style="vertical-align: -5px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=x^2.</math>
 |-
-|To do this, we need to solve &nbsp;<math>x=x^2.</math>
+|To do this, we need to solve &nbsp;<math style="vertical-align: 0px">x=x^2.</math>
 |-
 |Moving all the terms on one side of the equation, we get
@@ Line 33: / Line 33: @@
 \end{array}</math>
 |-
-|Hence, these two curves intersect at &nbsp;<math>x=0</math>&nbsp; and &nbsp;<math>x=1.</math>
+|Hence, these two curves intersect at &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=1.</math>
 |-
-|So, we are interested in the region between &nbsp;<math>x=0</math>&nbsp; and &nbsp;<math>x=1.</math>
+|So, we are interested in the region between &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=1.</math>
 |}
@@ Line 43: / Line 43: @@
 |We use the washer method to calculate this volume.
 |-
-|The outer radius is &nbsp;<math>r_{\text{outer}}=2-x^2</math>&nbsp; and
+|The outer radius is &nbsp;<math style="vertical-align: -4px">r_{\text{outer}}=2-x^2</math>&nbsp; and
 |-
-|the inner radius is &nbsp;<math>r_{\text{inner}}=2-x.</math>&nbsp;
+|the inner radius is &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math>&nbsp;
 |-
 |Therefore, the volume of the solid is

@@ Line 4: / Line 4: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' You can find the intersection points of two functions, say &nbsp; <math style="vertical-align: -5px">f(x),g(x),</math>
 |-
+|
 |-
+|'''2.''' The volume of a solid obtained by rotating an area around the &nbsp;<math style="vertical-align: -4px">x</math>-axis using the washer method is given by
 |-
+|
 |}
@@ Line 19: / Line 21: @@
 !Step 1: &nbsp;
 |-
+|First, we need to find the intersection points of &nbsp;<math>y=x</math>&nbsp; and &nbsp;<math>y=x^2.</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
+|Hence, these two curves intersect at &nbsp;<math>x=0</math>&nbsp; and &nbsp;<math>x=1.</math>
 |-
+|So, we are interested in the region between &nbsp;<math>x=0</math>&nbsp; and &nbsp;<math>x=1.</math>
 |}
@@ Line 31: / Line 41: @@
 !Step 2: &nbsp;
 |-
+|We use the washer method to calculate this volume.
 |-
+|Therefore, the volume of the solid is
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 44: / Line 78: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{8\pi}{15}</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves <math>y=x</math> and <math>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 &nbsp;<math style="vertical-align: -4px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; about the line &nbsp;<math>y=2.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 15: / Line 15: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 42: / Line 40: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 03:10, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">~~Consider~~ the region bounded by the ~~following two functions:~~	+	<span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves <math>y=x</math> and <math>y=x^2</math> about the line <math>y=2.</math>
−	~~::::::::<span class="exam">~~ <math ~~style="vertical-align: -5px"~~>y=~~2(-~~x~~^2+9)~~</math> and <math ~~style="vertical-align: -4px"~~>y=0</math>.
−
−	<~~span class="exam"~~>~~a) Using the lower sum with three rectangles having equal width, approximate the area.~~
−
−	~~<span class~~=~~"exam">b) Using the upper sum with three rectangles having equal width, approximate the area~~.
−
−	<~~span class="exam"~~>~~c) Find the actual area of the region.~~

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"