Question

Line integral

Answer 1

C is the arc of the curve x = y^3 from (-1,1) to (1,1) how do I parametrize this

Answer 2

\[\int\limits_C e^x dx\] @ganeshie8

Answer 3

y=t
x=t^3

will do right ?

Answer 4

Yeah, but how do you know that's what t is?

Answer 5

y=sint
x=sin^3t

works too

Answer 6

but you don't wanto deal with sin and cos so pick the simplest possible parameterization

Answer 7

Yeah I can see the result come out, but why do we let x = t^3 and y = t

Answer 8

we may choose anything we wish
the line integral wont depend on the parameterization

Answer 9

y=y
x=y^3

works too

Answer 10

Ah ok :), thanks

Answer 11

is there a typo

(-1,1) to (1,1)

because (-1, 1) is not on the given curve x=y^3

Answer 12

Oh yes, that should be (-1,-1)

Answer 13

\(y=t\)
\(x=t^3 \)
\(t : -1\longrightarrow 1\)

Answer 14

Yup, that's what I got

Answer 15

Did you get result \[e - \frac{ 1 }{ e }\]

Answer 16

Yup, I drew the same thing!

Answer 17

http://www.wolframalpha.com/input/?i=%5Cint_%28-1%29%5E1+e%5E%28y%5E3%29*3y%5E2+dy

Answer 18

I realized we just have to integrate e^x from -1 to 1, since we get \[\int\limits e^{y^3} 3y^2 dy = \int\limits e^x dx\]

Answer 19

once you setup the line integral,
it is just an integral... we can use all kinds of integration tricks..

Answer 20

:D

Answer 21

standalone \(\int_C f(x) dx\) makes no sense. there is no useful interpretation of this line integral involving only dx or dy

i think you're doing this as a practice to evaluate line integrals of vector fields

Answer 22

Yes, you're right! :P