# Vector field

• Apr 27th 2009, 08:03 AM
Apprentice123
Vector field
Calculate the circulation of the vector field $F(x,y) = (x,xy)$ around the closed curve $x^2+y^2 = 25$ situated above the axis $OX$
• Apr 27th 2009, 11:13 AM
Calculus26
we have talked before if F is a force field the line integral is the Work done

If F is a velocity field then the line integral is the circulation

so just compute the line integral of F over C

Her C consists of 2 parts the circle and the line segment along the x axis from -5 to 5

But of course you have Green's thm in your tool box as well
• Apr 27th 2009, 11:42 AM
Apprentice123
Quote:

Originally Posted by Calculus26
we have talked before if F is a force field the line integral is the Work done

If F is a velocity field then the line integral is the circulation

so just compute the line integral of F over C

Her C consists of 2 parts the circle and the line segment along the x axis from -5 to 5

But of course you have Green's thm in your tool box as well

Above the axis ox:

$\int_0^{ \pi} (-costsent + cos^2tsent)dt = - \frac{2}{3}$

Correct ?
• Apr 27th 2009, 12:11 PM
Calculus26
Not quite

x= 5cos(t) y = 5sin(t)

dr/dt = -5*sin(t) +5*cos(t)

F = 5cos(t)i +25cos(t)sin(t)

go from here
• Apr 27th 2009, 12:20 PM
Apprentice123
Quote:

Originally Posted by Calculus26
Not quite

x= 5cos(t) y = 5sin(t)

dr/dt = -5*sin(t) +5*cos(t)

F = 5cos(t)i +25cos(t)sin(t)

go from here

$\int_0^{ \pi} (-25costsint + 125 cos^2tsint)dt = \frac{250}{3}$

It is correct ?
• Apr 27th 2009, 12:28 PM
Calculus26
Ok--now you are correct----Again I've mentioned this before--slow down