# vector field

• Nov 9th 2009, 08:15 PM
acosta0809
vector field
The question:
For the vector field = -y + x, find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point $$(6)/sqrt(2), (6)/sqrt(2)$$. Give an exact answer.

i think from x=0 to x=6 it is 0 but how do i get the second part? help please
• Nov 10th 2009, 03:33 AM
HallsofIvy
Quote:

Originally Posted by acosta0809
The question:
For the vector field = -y + x,

This is not a vector field. Did you mean $-y\vec{i}+ x\vec{j}$?

Quote:

find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point $$(6)/sqrt(2), (6)/sqrt(2)$$. Give an exact answer.

i think from x=0 to x=6 it is 0 but how do i get the second part? help please
Assuming the the vector field to be integrated is indeed $-y\vec{i}+ x\vec{j}$, on the path from (0,0) to (6, 0) we can take x= t ( $0\le t\le 6$) and y= 0. Then dx= dt and y= 0dt so \math]\int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy}= \int (0dx+ x(0))= 0[/tex].

Write the circle $x^2+ y^2= 36$ as x= 6 cos(t), y= 6 sin(t) (that's a standard parameterization for a circle) so that dx= -6 sin(t)dt and dy= 6 cos(t) dt. The integral becomes $\int (-6 sin(t)\vec{i}+ 6 cos(t)\vec{j})\cdot(-6 sin(t)dt\vec{i}+ 6 cos(t)dt\vec{j})$ $= \int(36sin^2(t)dt+ 36cos^2(t))dt= 36\int dt$.

Or course, if x= 6 cos(t) then x= $\frac{6}{\sqrt{2}}$ when $cos(t)= 1/\sqrt{2}$ and that happens when $t= \pi/4$. The integral is from t= 0 to $t= \pi/4$.

(Blast! I was hoping I could get back and correct my silly mistake [I mixed up x and y] before anyone had looke at it!)
• Nov 10th 2009, 04:02 AM
tonio
Quote:

Originally Posted by HallsofIvy
This is not a vector field. Did you mean $-y\vec{i}+ x\vec{j}$?

Assuming the the vector field to be integrated is indeed $-y\vec{i}+ x\vec{j}$, on the path from (0,0) to (6, 0) we can take x= t ( $0\le t\le 6$) and y= 0. Then dx= dt and y= 0 dt so $\int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy)= \int (0dx+ x(0))= 0$.

Write the circle $x^2+ y^2= 36$ as x= 6 cos(t), y= 6 sin(t) (that's a standard parameterization for a circle) so that dx= -6 sin(t)dt and dy= 6 cos(t) dt. The integral becomes $\int (6 cos(t)\vec{i}+ 6 sin(t)\vec{j})\cdot(-6 sin(t)dt\vec{i}+ 6 cos(t)dt\vec{j})$ $= \int(-36sin(t)cos(t)dt+ 36cos(t)sin(t))dt= 0$!

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