anonymous
  • anonymous
So this question had me stumped on my test earlier. Find the surface area: y=sqrt(1-(x^2/4)) where 0 is less than or equal to x less than or equal to 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[y=\sqrt{1-x^2/4}\]
anonymous
  • anonymous
\[0\le x \le2\]
TuringTest
  • TuringTest
surface area? is this a revolution?

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anonymous
  • anonymous
And the formula is \[\int\limits_{a}^{b}2\Pi yds\]
anonymous
  • anonymous
\[ds=\sqrt{1+(dy/dx)^2}\]
TuringTest
  • TuringTest
so you are revolving around the x-axis?
anonymous
  • anonymous
Yesh, sorry for not answering
anonymous
  • anonymous
about the x-axis
TuringTest
  • TuringTest
Ok here's an idea...
anonymous
  • anonymous
Any ideas are much appreciated!
TuringTest
  • TuringTest
\[y=\sqrt{1-\frac{x^2}4}\implies y^2=1-\frac{x^2}4\]using implicit differentiation we get\[2yy'=-\frac x2\implies y'=-{x\over4y}\]\[2\pi\int yds=2\pi\int y\sqrt{1+{x^2\over16y^2}}dx=2\pi\int_{0}^{2}\sqrt{y^2+{x^2\over 16}}dx\]I'll continue, but let you digest in the meantime...
anonymous
  • anonymous
Sounds good, willing to wait
anonymous
  • anonymous
Wait, is not ok to leave it in terms of x since we are rotating about the x-axis?? and by "y" in the formula, it means whatever y=. Just to clear up some confusion.
TuringTest
  • TuringTest
but this is where the trick of implicit differentiation pays off: we will now sub in that expression for y into the formula above:\[2\pi\int_{0}^{2}\sqrt{1-{x^2\over4}+{x^2\over16}}dx=2\pi\int_{0}^{2}\sqrt{1-{3x^2\over16}}dx\]and now we have a trig sub... I hope
TuringTest
  • TuringTest
more specifically we subbed in for \(y^2\)
anonymous
  • anonymous
Haha oh ok. I was thinking I would need a trig substitution, but radical was not very pretty looking
TuringTest
  • TuringTest
yeah, but waiting to sub in for y made it a lot easier
anonymous
  • anonymous
cause my radical looked very similar but my trig substitution was not making any sense. This is what i had on my test, which I know is wrong. \[\pi/2\int\limits_{0}^{2} \sqrt{-3x^2+16}\]
TuringTest
  • TuringTest
it is the same as mine, but with pi/2 for some reason I don't get...
anonymous
  • anonymous
Well I had 16 on bottom, so i split the radical up into to with numerator squared over the denominator squared. And the sqrt(16) is 4 so I took out the constant of 1/4 and that times 2pi=pi/2. But my trig sub. was not working out to nicely, so I figured I messed up somewhere.
TuringTest
  • TuringTest
oh yeah, my algebra was off you are right about the pi/2 that said, we have the same thing\[\sqrt{-3x^2+16}=\sqrt{16-3x^3}=\frac14\sqrt{1-{3x^2\over16}}\]so try the sub\[\frac4{\sqrt3}\sin\theta=x\]
anonymous
  • anonymous
ohhhh that is where I messed up, all i did was \[1/\sqrt{3}\] \[\sin \theta\]
anonymous
  • anonymous
well those are suppose to be together. haha
TuringTest
  • TuringTest
gotta get rid of that 16 ;) I gotta go, glad I could help!
anonymous
  • anonymous
Yes, thank you very much!!!

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