anonymous
  • anonymous
Find the surface area that results when the curve is rotated around the x-axis: x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
????
myininaya
  • myininaya
use int(2*pi*y*sqrt((x')^2+(y')^2),t)
myininaya
  • myininaya
int(2*pi*t*sqrt(4t^2+1),t)

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myininaya
  • myininaya
let u=4t^2+1 so du=8tdt so we have int(2/8*u^(1/2),u) that should help you
anonymous
  • anonymous
ok i'm tryin it now
amistre64
  • amistre64
would this be equivalent to: x = y^2 +1 ?
anonymous
  • anonymous
im still workin on it ha
amistre64
  • amistre64
i spose the line integral dont care if its parametric or not eh :)
anonymous
  • anonymous
what im confused ha
amistre64
  • amistre64
the line integral;\[\int\limits_{} \sqrt{f'(x) + g'(x)} dx\]
anonymous
  • anonymous
i dont know what exactly you're asking but we're doing parametric stuff now?
anonymous
  • anonymous
haha i'm so confused with this stuff right now sorry
amistre64
  • amistre64
the line integral is primed for parametric equations; you insert the derivatives of your parameter equations and run with it :)
anonymous
  • anonymous
so i should find the derivative of both and plug them into the integral you put earlier instead of what i just did?
amistre64
  • amistre64
dunno what you just did :) but yeah. you find the length of the line, and spin it to find the surface area...
amistre64
  • amistre64
\[2\pi \int\limits_{0}^{2}\sqrt{f'(t) + g'(t)} dt\]
amistre64
  • amistre64
if im wrong, im sure someone will let me know :)
amistre64
  • amistre64
x = t ^(2) + 1 , y = t x' = 2t ; y' = 1 fill em in
amistre64
  • amistre64
and I was wrong...those whoud be "squared" lol
amistre64
  • amistre64
\[2\pi \int\limits_{0}^{2}\sqrt{[f'(x)]^{2} + [g'(x)]^2}\]
anonymous
  • anonymous
ok so i ended up with pi/4(2u^(3/2)/3) with the limits from 1 to 17 bc i plugged in 0 and 2 into u = 4t^2 +1
anonymous
  • anonymous
and just going to plug in those limits to find out the answer is that right?
amistre64
  • amistre64
[S] (4t^2 + 1)^(1/2) dt how did you go about integrating the square root? did you use a trig substitutiion? or a "u" substitution?
anonymous
  • anonymous
u=4t^2 +1
anonymous
  • anonymous
so du = 8tdt
amistre64
  • amistre64
and how does that integrate?...good, and how did you get rid of the "t" from that?
anonymous
  • anonymous
ohhhhh i guess i messed that up
anonymous
  • anonymous
ahh now im confused again
amistre64
  • amistre64
\[\int\limits_{}8t (\sqrt{u}) du =?\]
amistre64
  • amistre64
itd be in the denominator, but still....its there lol
amistre64
  • amistre64
trig identities are what you can use for substitution in this
amistre64
  • amistre64
\[\sqrt{4x^2 + 1}\] tan^2 + 1 = sec^2 right?
amistre64
  • amistre64
say u = 2x tan(x) u^2 = 4x^2 tan^2....plays a role in this someplace :)
anonymous
  • anonymous
im not gettin it ha
amistre64
  • amistre64
you pretty much want to make it looke like this:\[\sqrt{\tan^2(x) + 1} = \sec(x)\]
amistre64
  • amistre64
so that tan(x) = 2x
amistre64
  • amistre64
then you integrate sec(x) dx and turn the substitution around
anonymous
  • anonymous
so would it be (2tan^2(x) +1)^1/2
anonymous
  • anonymous
when i start the trig sub or what?
amistre64
  • amistre64
no, then you cant "factor" out that "2" in front of the tan... just my thoughts :)
anonymous
  • anonymous
ahhh ive been workin at this too long i cant think straight now im forgetting how to do trig sub stuff
amistre64
  • amistre64
4x^2 needs to equal tan^2 2x = tan then
amistre64
  • amistre64
we draw a triangle with the corresponding parts so that one leg = 2x; the other leg = 1 and the hypotenuse = whatever it equals :)
amistre64
  • amistre64
that way we can convert between our stuff in the end right?
anonymous
  • anonymous
ya i mean if we did the triangle it would be sqrt of 2x^2 + 1^1
anonymous
  • anonymous
ok so heres where im at
amistre64
  • amistre64
does this help?
amistre64
  • amistre64
1 Attachment
anonymous
  • anonymous
2 pi (ln(sec(t) + tan(t)) and whatever the limits are
anonymous
  • anonymous
does that seem correct?
amistre64
  • amistre64
that looks good, just need to replace sec and tan with their appropriate values
anonymous
  • anonymous
ok so tan is 2x right?
amistre64
  • amistre64
correct.... or 2t...whatever you wanna call it :)
anonymous
  • anonymous
and sec is adj/opp? so 1/2x
anonymous
  • anonymous
?
amistre64
  • amistre64
sec = hyp/adj
amistre64
  • amistre64
its cos upside down
anonymous
  • anonymous
ya thats right
amistre64
  • amistre64
\[1/\sqrt{4t^2 +1}\]
anonymous
  • anonymous
if its hyp/adj shouldnt it just be \[\sqrt{4t^2+1}\]
amistre64
  • amistre64
.... i spose if you want to be absolutley technical lol
amistre64
  • amistre64
[ln(sqrt(17) + 4)] - [ln(1)]
amistre64
  • amistre64
ummm....4(4) = 16 + 1 = 17 right?
anonymous
  • anonymous
ya
amistre64
  • amistre64
and the other side is sqrt(4(0) + 1) + 2(0) = ln(1)
amistre64
  • amistre64
i hope you got an answer key, cause id love to know if were right on this :)
amistre64
  • amistre64
2.09 is what I get....

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