Find the surface area that results when the curve is rotated around the x-axis:
x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .

- anonymous

- jamiebookeater

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- anonymous

????

- myininaya

use int(2*pi*y*sqrt((x')^2+(y')^2),t)

- myininaya

int(2*pi*t*sqrt(4t^2+1),t)

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## More answers

- myininaya

let u=4t^2+1 so du=8tdt so we have int(2/8*u^(1/2),u) that should help you

- anonymous

ok i'm tryin it now

- amistre64

would this be equivalent to:
x = y^2 +1 ?

- anonymous

im still workin on it ha

- amistre64

i spose the line integral dont care if its parametric or not eh :)

- anonymous

what im confused ha

- amistre64

the line integral;\[\int\limits_{} \sqrt{f'(x) + g'(x)} dx\]

- anonymous

i dont know what exactly you're asking but we're doing parametric stuff now?

- anonymous

haha i'm so confused with this stuff right now sorry

- amistre64

the line integral is primed for parametric equations; you insert the derivatives of your parameter equations and run with it :)

- 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

dunno what you just did :) but yeah.
you find the length of the line, and spin it to find the surface area...

- amistre64

\[2\pi \int\limits_{0}^{2}\sqrt{f'(t) + g'(t)} dt\]

- amistre64

if im wrong, im sure someone will let me know :)

- amistre64

x = t ^(2) + 1 , y = t
x' = 2t ; y' = 1 fill em in

- amistre64

and I was wrong...those whoud be "squared" lol

- amistre64

\[2\pi \int\limits_{0}^{2}\sqrt{[f'(x)]^{2} + [g'(x)]^2}\]

- 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

and just going to plug in those limits to find out the answer is that right?

- 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

u=4t^2 +1

- anonymous

so du = 8tdt

- amistre64

and how does that integrate?...good, and how did you get rid of the "t" from that?

- anonymous

ohhhhh i guess i messed that up

- anonymous

ahh now im confused again

- amistre64

\[\int\limits_{}8t (\sqrt{u}) du =?\]

- amistre64

itd be in the denominator, but still....its there lol

- amistre64

trig identities are what you can use for substitution in this

- amistre64

\[\sqrt{4x^2 + 1}\]
tan^2 + 1 = sec^2 right?

- amistre64

say u = 2x tan(x)
u^2 = 4x^2 tan^2....plays a role in this someplace :)

- anonymous

im not gettin it ha

- amistre64

you pretty much want to make it looke like this:\[\sqrt{\tan^2(x) + 1} = \sec(x)\]

- amistre64

so that tan(x) = 2x

- amistre64

then you integrate sec(x) dx and turn the substitution around

- anonymous

so would it be (2tan^2(x) +1)^1/2

- anonymous

when i start the trig sub or what?

- amistre64

no, then you cant "factor" out that "2" in front of the tan... just my thoughts :)

- anonymous

ahhh ive been workin at this too long i cant think straight now im forgetting how to do trig sub stuff

- amistre64

4x^2 needs to equal tan^2
2x = tan then

- 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

that way we can convert between our stuff in the end right?

- anonymous

ya i mean if we did the triangle it would be sqrt of 2x^2 + 1^1

- anonymous

ok so heres where im at

- amistre64

does this help?

- amistre64

##### 1 Attachment

- anonymous

2 pi (ln(sec(t) + tan(t)) and whatever the limits are

- anonymous

does that seem correct?

- amistre64

that looks good, just need to replace sec and tan with their appropriate values

- anonymous

ok so tan is 2x right?

- amistre64

correct.... or 2t...whatever you wanna call it :)

- anonymous

and sec is adj/opp? so 1/2x

- anonymous

?

- amistre64

sec = hyp/adj

- amistre64

its cos upside down

- anonymous

ya thats right

- amistre64

\[1/\sqrt{4t^2 +1}\]

- anonymous

if its hyp/adj shouldnt it just be \[\sqrt{4t^2+1}\]

- amistre64

.... i spose if you want to be absolutley technical lol

- amistre64

[ln(sqrt(17) + 4)] - [ln(1)]

- amistre64

ummm....4(4) = 16 + 1 = 17 right?

- anonymous

ya

- amistre64

and the other side is sqrt(4(0) + 1) + 2(0) = ln(1)

- amistre64

i hope you got an answer key, cause id love to know if were right on this :)

- amistre64

2.09 is what I get....

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