Finding volume using integrals

  • Thread starter donjt81
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  • #1
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So here is the question

find the volume of the solid generated by revolving the region bounded by the curve y = sqrt(16 - x^2) and the line y = 0 about the x axis.

this is how I solved it

[tex] \int_{0}^{4} \Pi (16 - y^{2}) \; dy [/tex]
[tex] \Pi \int_{0}^{4} (16 - y^{2}) \; dy [/tex]

pi(16*y - y^3/3) from 0 to 4
pi(16*4 - 64/3 - 0)

so the answer I got is (128 pi)/3

is this correct.
 

Answers and Replies

  • #2
radou
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Unless I'm missing something, you should multiply your whole integral with 2, since your function is defined from -4 to 4 and symmetric.
 
  • #3
arildno
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Well, why haven't you used -4 as your lower limit?
Note that this would double the volume of your object.
 
  • #4
HallsofIvy
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Another check is to recognize that [itex]y= \sqrt{16- x^2}[/itex] is the upper half of a circle and so the figure formed is a sphere of radius 4. Its volume is [itex](4/3)\pi (4)^3[/itex], twice your answer.
 
  • #5
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thanks guys i completely missed that. you are right it should be -4 to 4... other than that does everything else look ok.
 

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