Yttrium
  • Yttrium
Find the surface area of the solid of revolution generated by revolving about the x-axis the arc of y^2 + 4x = 2lny from y = 1 to y = 3.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Yttrium
  • Yttrium
answer this please.
anonymous
  • anonymous
\[\int\limits_{}^{}2*\Pi*y*\sqrt{dy^2+dx^2}\]
anonymous
  • anonymous
surface area integral

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More answers

Yttrium
  • Yttrium
I'm done with those things, but the part making it difficult for me is the integration part. the sqrt(y^4 + y^2 + 1)
anonymous
  • anonymous
\[\int\limits_{1}^{3}2*\pi*y*\sqrt{1+(dx/dy)^2}dy\]
Yttrium
  • Yttrium
Yes. That's make me overthink. Can you help me?
anonymous
  • anonymous
@dape this is not a surface area integral.
anonymous
  • anonymous
can you find dx/dy ? 2ydy +4dx=2dy/y (2y-2/y)dy = -4dx dx/dy = 1/(2y) - y/2 (dx/dy)^2 = 1/(4y^2) - 1/2 + y^2/4 1 + (dx/dy)^2 = 1/(4y^2) + 0.5 + y^2/4 so we need to integrate 2pi*sqrt(0.25 + 0.5y^2 + y^4/4) pi * sqrt(1+y^2 + y^4) as you got @Yttrium
Yttrium
  • Yttrium
So what shall I do next? :D
dape
  • dape
@Coolsector that's gonna be one hell of an integral.
anonymous
  • anonymous
well lol what can i do
dape
  • dape
You are gonna have to use elliptic functions to solve it.
anonymous
  • anonymous
but @dape the integral that you wrote wont give the surface area
Yttrium
  • Yttrium
yeah i agree @Coolsector
dape
  • dape
Something seems off, I'm getting a negative answer.
anonymous
  • anonymous
maybe we need another approach
Yttrium
  • Yttrium
btw, how did you get that kind of high smarscore?
anonymous
  • anonymous
what is it ? lol
anonymous
  • anonymous
@dape , i know that it is tempting to write the differential dx or dy but we really need to work with ds (the length) http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx look at surface area formulas there
anonymous
  • anonymous
the hell!! i wrote : so we need to integrate 2pi*sqrt(0.25 + 0.5y^2 + y^4/4) pi * sqrt(1+y^2 + y^4) but it should be 2pi*sqrt(0.25 + 0.5y^2 + y^4/4) pi * sqrt(1+2y^2 + y^4)
anonymous
  • anonymous
so pi sqrt((y^2+1)^2) pi(y^2+1) !!!
anonymous
  • anonymous
that is easy to integrate.. just a calculation mistake
anonymous
  • anonymous
so the surface area is pi * (y^3/3 + y) from 1 to 3 which is 28 and 2/3 times pi
anonymous
  • anonymous
|dw:1378287223929:dw| first we need to write the formula, since we have the notation about the x-axis, surface area is: on the board then we pick the 2nd definition of ds since we are given 1 <= y <= 3 as the bounds.
anonymous
  • anonymous
|dw:1378287731983:dw|
anonymous
  • anonymous
@ⒶArchie☁✪ i found the problem look at my last responses no point in doing it all over again.
anonymous
  • anonymous
so for our problem we can write like that on the board, then....
anonymous
  • anonymous
well I'm just explaining it to Yttrium, so he is all confirmed.
anonymous
  • anonymous
@Yttrium, would you like me to continue with this problem or you seem alright with it now?
dape
  • dape
Yeah, you're right, we have to use the differential curve length, not just dy.
anonymous
  • anonymous
I know how to solve this, but since Yttrium is not available at the moment. I'll just hold...
Yttrium
  • Yttrium
I found the error. And it easy, indeed. :D
anonymous
  • anonymous
:)

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