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anonymous
 5 years ago
Find the surface area that results when the curve is rotated around the xaxis:
x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .
anonymous
 5 years ago
Find the surface area that results when the curve is rotated around the xaxis: x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0use int(2*pi*y*sqrt((x')^2+(y')^2),t)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0int(2*pi*t*sqrt(4t^2+1),t)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0let u=4t^2+1 so du=8tdt so we have int(2/8*u^(1/2),u) that should help you

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0would this be equivalent to: x = y^2 +1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im still workin on it ha

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i spose the line integral dont care if its parametric or not eh :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the line integral;\[\int\limits_{} \sqrt{f'(x) + g'(x)} dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know what exactly you're asking but we're doing parametric stuff now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha i'm so confused with this stuff right now sorry

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the line integral is primed for parametric equations; you insert the derivatives of your parameter equations and run with it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i should find the derivative of both and plug them into the integral you put earlier instead of what i just did?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dunno what you just did :) but yeah. you find the length of the line, and spin it to find the surface area...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits_{0}^{2}\sqrt{f'(t) + g'(t)} dt\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if im wrong, im sure someone will let me know :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = t ^(2) + 1 , y = t x' = 2t ; y' = 1 fill em in

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and I was wrong...those whoud be "squared" lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits_{0}^{2}\sqrt{[f'(x)]^{2} + [g'(x)]^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0and just going to plug in those limits to find out the answer is that right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and how does that integrate?...good, and how did you get rid of the "t" from that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhh i guess i messed that up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh now im confused again

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}8t (\sqrt{u}) du =?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0itd be in the denominator, but still....its there lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0trig identities are what you can use for substitution in this

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{4x^2 + 1}\] tan^2 + 1 = sec^2 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0say u = 2x tan(x) u^2 = 4x^2 tan^2....plays a role in this someplace :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you pretty much want to make it looke like this:\[\sqrt{\tan^2(x) + 1} = \sec(x)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then you integrate sec(x) dx and turn the substitution around

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would it be (2tan^2(x) +1)^1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i start the trig sub or what?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no, then you cant "factor" out that "2" in front of the tan... just my thoughts :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh ive been workin at this too long i cant think straight now im forgetting how to do trig sub stuff

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04x^2 needs to equal tan^2 2x = tan then

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we draw a triangle with the corresponding parts so that one leg = 2x; the other leg = 1 and the hypotenuse = whatever it equals :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that way we can convert between our stuff in the end right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya i mean if we did the triangle it would be sqrt of 2x^2 + 1^1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so heres where im at

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02 pi (ln(sec(t) + tan(t)) and whatever the limits are

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that seem correct?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that looks good, just need to replace sec and tan with their appropriate values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so tan is 2x right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0correct.... or 2t...whatever you wanna call it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and sec is adj/opp? so 1/2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if its hyp/adj shouldnt it just be \[\sqrt{4t^2+1}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.... i spose if you want to be absolutley technical lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[ln(sqrt(17) + 4)]  [ln(1)]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ummm....4(4) = 16 + 1 = 17 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and the other side is sqrt(4(0) + 1) + 2(0) = ln(1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i hope you got an answer key, cause id love to know if were right on this :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02.09 is what I get....
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