Question

evaluate the line integral, where C is the given curve:

Answer 1

\[\int\limits_{}^{}y ^{3}dS\] C: x= t^3, y = t, 0<t<2

Answer 2

i sep getting (1/36)(145^(3/2) - 1) and it should be (1/54)(145^(3/2) - 1)

Answer 3

\[ds=\sqrt{(3t^2)^2+1}=\sqrt{9t^4+1}\]\[\int_{0}^{2}t^3\sqrt{9t^4+1}dt\]This should be the right setup I think

Answer 4

@allyfranken yes, this gives the right answer
let me know where you are having your trouble

Answer 5

it gives you 1/54? bc i et 1/36 when i do it

Answer 6

yeah i did the integral above and i got 1/36 instead of 1/54

Answer 7

wouldn't ds be ||r'(t)|| not ||r(t)|| though?

Answer 8

Looks like Turing was right after all.

Answer 9

http://www.wolframalpha.com/input/?i=integrate+from+0+to+2+t%5E3+sqrt%289t%5E4+%2B+1%29+dt

Answer 10

can you walk me through it bc i keep getting 1/36 instead of 1/54

Answer 11

you meant he integral??

Answer 12

yup

Answer 13

omg nvm i got it

Answer 14

let t^2 = 1/3 tan x

Answer 15

2t dt = 1/2 sec^2x dx

Answer 16

dt = 1/4 sec^2x/sqrt(tanx)

Answer 17

*edit* dt = 1/4 sec^2x sqrt(3)/sqrt(tanx) dx
y^3 = tan^3/2x /3sqrt(3)

y^3 dt = tan^(3/2)x /3sqrt(3) * 1/4 sec^2x sqrt(3)/sqrt(tanx) dx

= tan^(3/2)x /3sqrt(3) * 1/4 sec^2x sqrt(3)/sqrt(tanx) dx
= 1/12 * tanx sec^x dx

Answer 18

what have you got??

Answer 19

after putting that values in square root we get
1/12 * tanx sec^3x dx

Answer 20

seems like I got the same
http://www.wolframalpha.com/input/?i=integrate+1%2F12+tanx+sec%5E3x

Answer 21

@experimentX why did you not just use\[u=9t^4+1\]?
:P

Answer 22

lol ... didn't see that.

Answer 23

It could have made a lot simpler. since i've used that kind of method one or two times here on OS, i've not build insight.
Well, a step to remember.

Answer 24

just had to point it out
simplicity=elegance