Luigi0210
  • Luigi0210
help with volume
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Luigi0210
  • Luigi0210
Find the volume of the solid formed by rotating the ellipse x^2/a^2+y^2/b^2=1about the x-axis
primeralph
  • primeralph
just square and integrate?
anonymous
  • anonymous
i'd have to look up to refresh how to do it again but you should manipulate it to: \[y = \pm b \sqrt{1 - (x/a)^2}\] and treat it like a single variable function. easiest to do two times the positive half. if you would like to know about the cylinder method to get to the answer, let me know

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anonymous
  • anonymous
my bad. its around the x axis, so use the discs method
Luigi0210
  • Luigi0210
Could you walk me through this to the solution? Volume was never easy for me
anonymous
  • anonymous
do you get how i got the function? ill start drawing
Luigi0210
  • Luigi0210
Yea, I understood that part
primeralph
  • primeralph
.........there's an easy way.....
primeralph
  • primeralph
|dw:1370928893209:dw|
primeralph
  • primeralph
|dw:1370928986118:dw|
anonymous
  • anonymous
|dw:1370928683857:dw| |dw:1370928978711:dw| \[A_{disc} = \pi r^2 = \pi (f(x))^2\] Volume of disc = pi f(x)^2 dx I wrote the integral wrong the first time, ralph wrote it write though. Volume would be equal to the sum of all the infinitesimally of varying radii. According to definition of integral it can be represented as. \[V = \int\limits_{-a}^{a} \pi f(x)^2 dx = \pi \int\limits_{-a}^{a} b^2(1 - (x/a)^2)dx\] forget what i said earlier about taking two times the top. this entire method is the way to do it
primeralph
  • primeralph
eerrrr.....ain't that what I said?
anonymous
  • anonymous
you didn't explain why
anonymous
  • anonymous
**Volume would be equal to the sum of all the infinitesimally thin discs of varying radii.* corrected sentence
Luigi0210
  • Luigi0210
Makes sense. Thanks! @Euler271 @primeralph
primeralph
  • primeralph
errr. it's in the textbook ain't it?
primeralph
  • primeralph
@Euler271 Luigi is smart enough to figure stuff out
KenLJW
  • KenLJW
The limits appear to be wrong, integral always seem to be zero? If you solve for y and integrate from (-a, 0) to (a, 0) you always get 0 and should get area under the curve. The way around this integrate from (-a, 0) to (0, 0) and multiply by 2, this give area of top half. It seems to me your rotating therefore multiply by 2 pi. Please give your actual solutions so I can understand. An innocent bystander.

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