Question

Find the surface area that results when the curve is rotated around the x-axis:

x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .

Answer 1

????

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

ok i'm tryin it now

Answer 6

would this be equivalent to:

x = y^2 +1 ?

Answer 7

im still workin on it ha

Answer 8

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

Answer 9

what im confused ha

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

so i should find the derivative of both and plug them into the integral you put earlier instead of what i just did?

Answer 15

dunno what you just did :) but yeah.

you find the length of the line, and spin it to find the surface area...

Answer 16

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

Answer 17

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

Answer 18

x = t ^(2) + 1 , y = t

x' = 2t ; y' = 1 fill em in

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

[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?

Answer 24

u=4t^2 +1

Answer 25

so du = 8tdt

Answer 26

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

Answer 27

ohhhhh i guess i messed that up

Answer 28

ahh now im confused again

Answer 29

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

Answer 30

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

Answer 31

trig identities are what you can use for substitution in this

Answer 32

\[\sqrt{4x^2 + 1}\]

tan^2 + 1 = sec^2 right?

Answer 33

say u = 2x tan(x)

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

Answer 34

im not gettin it ha

Answer 35

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

Answer 36

so that tan(x) = 2x

Answer 37

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

Answer 38

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

Answer 39

when i start the trig sub or what?

Answer 40

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

Answer 41

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

Answer 42

4x^2 needs to equal tan^2

2x = tan then

Answer 43

we draw a triangle with the corresponding parts so that one leg = 2x; the other leg = 1 and the hypotenuse = whatever it equals :)

Answer 44

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

Answer 45

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

Answer 46

ok so heres where im at

Answer 47

does this help?

Answer 48

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

Answer 49

does that seem correct?

Answer 50

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

Answer 51

ok so tan is 2x right?

Answer 52

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

Answer 53

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

Answer 54

?

Answer 55

sec = hyp/adj

Answer 56

its cos upside down

Answer 57

ya thats right

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

ya

Answer 64

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

Answer 65

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

Answer 66

2.09 is what I get....