1. ## Integral of line

Calculate the integral of line $\int_c F.dr$, where $F(x,y) = \sqrt{x^2+y^2}$ and c is the circumference $x^2+y^2=1$

How do I calculate integral of line ?

2. Originally Posted by Apprentice123
Calculate the integral of line $\int_c F.dr$, where $F(x,y) = \sqrt{x^2+y^2}$ and c is the circumference $x^2+y^2=1$

How do I calculate integral of line ?

Are you sure it's $F\cdot r$ with your given $F$?

3. Yes the answer is $2 \pi$

4. Apprentice

In English we call it a line integral

The question only makes sense If F is a vector field

Do you mean F = r = xi +yj ?

In which case use x = cos(t) y = sin(t) to parameterize the curve

r = cos(t) i +sin(t) j dr/dt = -sin(t) i +cos(t) j

On the is curve F = cos(t) i +sin(t) j

F*dr/dt = 0

Therefor the line integral is 0

If you want check out the line integral page on my website for the general method

Line Integrals

5. Originally Posted by Apprentice123
Yes the answer is $2 \pi$
But ${\bf F} \cdot d {\bf r}$ only makes sense if both are vectors!

6. dr is understood to be the vector dx i +dy j

Recall For line integrals to actually calculate we use

integral of F(x(t),y(t)) *dr/dt as t varies from a to b

F*dr is just notation rarely used just like the notation integral(fdx+gdy)

where F = f i + g j

We use integral of F(x(t),y(t)) *dr/dt

7. This is correct ?

$\int_0^{2 \pi} \sqrt{(cos(t))^2 + (sin(t))^2}.(-sin(t),cos(t))dt$

8. what you have makes no sense --F must be a vector field!!!!

So what is F ?

9. Sorry, what are the steps to resolve line integral ?

10. Is it possible that what you want is $\int_c F\,dx$, $\int_c F\,dy$ or $\int_c F\,ds$?

11. You start with a vector field F = f(x,y) i + g(x,y) j

parameterize the curve r = x(t) i +y(t) j

Then then the line integral is the integral of F(x(t),y(t))* dr/dt Where * is the dot product

Integrate F(x(t),y(t))* dr/dt from a to b

12. Again check out

Line Integrals for the development of line integrals

13. This line integral is:

$\int_0^{2 \pi} \sqrt{cos(t)^2}(-sen(t))dt + \int_0^{2 \pi} \sqrt{sen(t)^2}(cos(t))dt$

$= \frac{-1}{2} \frac{cos(t)^3}{3}|_0^{2 \pi} + \frac{-1}{2} \frac{sen(t)^3}{3} |_0^{2 \pi} = 0$

????

14. Apprentice

What is the vector field F you start with? ---If we don't know F we can't proceed!!!!!!!!!!!!!!!

F= (x^2 +y^2)^(1/2) is not a vector field

Look at the problem again there must be a vector field F !!!!!!!!!!!!!!

15. This problem i find in internet but not correct.

This problem is the book:

1) C is the curve represented by the equations:
$x=2t; y=3t^2; (0 \leq t \leq 1)$

Calculate the line integral along C

a) $\int_c (x-y)ds$
b) $\int_c (x-y)dx$
c) $\int_c (x-y)dy$

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