Question

et f=, evaluate the line integral below where c is the parabola on the xy plane: y =x^2, 0<=x<=1

Answer 1

are you a teacher or how do you know all this so well?

Answer 2

line integral along curve \(C\) : \[\int_C x dx + (2x+3y)dy\]

Answer 3

parameterize and evaluate

Answer 4

given curve : \(y=x^2\)

see if below parameterization works :
\(x = t\)
\(y = t^2\)

Answer 5

yes it does

\(x = t \implies dx= dt\)
\(y = t^2 \implies dy = 2t ~dt \)

Answer 6

plug these in ur line integral and evaluate

Answer 7

\[\int_C x dx + (2x+3y)dy = \int_0^1 t ~dt + (2t + 3t^2) 2tdt\]

Answer 8

you can evaluate that

Answer 9

10/3?

Answer 10

Yep !
http://www.wolframalpha.com/input/?i=+%5Cint_0%5E1+++t+%2B++2t%282t+%2B+3t%5E2%29dt

Answer 11

find the flux of the vector field f=x,y,z> acros the slanted surface of the cone z^2=x^2+y^2 for 0<=z<=1: normal vector points in the positive z direction

Answer 12

you do double integral of f*nds

Answer 13

I need to revise flux, and yes its just a double integral

Answer 14

its F n dS

Answer 15

earlier, you found normal vector for cone : \(\left< \dfrac{x}{z}, \dfrac{y}{z} , -1 \right>\)

Answer 16

yeah we use that vector but since they want positive z
dont you flip your signs

Answer 17

you mean : \(\left< -\dfrac{x}{z}, -\dfrac{y}{z} , 1 \right>\)

like this ?

Answer 18

i think so

Answer 19

or not sure

Answer 20

\(\overrightarrow{F} \bullet \hat{n} = \left<x,y,z\right> \bullet \dfrac{\left< -\dfrac{x}{z}, -\dfrac{y}{z} , 1 \right>}{\sqrt{2}} \)

Answer 21

i think you're right. see if above looks okay

Answer 22

\(\sqrt{2}\) is there becoz you want a unit vector, i have just divided the normal vector by its magnitude..

Answer 23

how did you get sqrt of 2

Answer 24

\(\sqrt{2}\) is the magnitude of "normal vector"

Answer 25

\( \Bigg|\left< -\dfrac{x}{z}, -\dfrac{y}{z} , 1 \right>\Bigg| = \sqrt{\dfrac{x^2}{z^2} + \dfrac{y^2}{z^2} + 1^2} = \sqrt{\dfrac{x^2+y^2}{z^2} + 1} = \sqrt{\dfrac{z^2}{z^2} + 1} = \sqrt{2}\)

Answer 26

ok

Answer 27

i dont think it works i think we have to use the divergence theorem

Answer 28

we want the flux only through the slant surface right ?