Difference between revisions of "009B Sample Final 1, Problem 5"
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!Step 1: | !Step 1: | ||
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− | | | + | |We proceed using cylindrical shells. The radius of the shells is given by <math>r=x</math>. |
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− | | | + | |The height of the shells is given by <math>h=e^x-ex</math>. |
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!Step 2: | !Step 2: | ||
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− | | | + | |So, the volume of the solid is <math>\int_0^1 2\pi x(e^x-ex)~dx</math>. |
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|'''(a)''' <math>(1,e)</math> (See (a) Step 1 for the graph) | |'''(a)''' <math>(1,e)</math> (See (a) Step 1 for the graph) | ||
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− | |'''(b)''' | + | |'''(b)''' <math>\int_0^1 2\pi x(e^x-ex)~dx</math> |
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|'''(c)''' | |'''(c)''' | ||
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[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
Revision as of 18:35, 4 February 2016
Consider the solid obtained by rotating the area bounded by the following three functions about the -axis:
- , , and .
a) Sketch the region bounded by the given three functions. Find the intersection point of the two functions:
and . (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.
Foundations: |
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Review volumes of revolutions |
Solution:
(a)
Step 1: |
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First, we sketch the region bounded by the three functions. |
Insert graph here. |
Step 2: |
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Setting the equations equal, we have . |
We get one intersection point, which is . |
This intersection point can be seen in the graph shown in Step 1. |
(b)
Step 1: |
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We proceed using cylindrical shells. The radius of the shells is given by . |
The height of the shells is given by . |
Step 2: |
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So, the volume of the solid is . |
(c)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) (See (a) Step 1 for the graph) |
(b) |
(c) |