Bend and Circulate

Calculus Level pending

Consider the following vector field:

F = ( F x , F y , F z ) = ( y , x , 0 ) \large{\vec{F} = (F_x,F_y,F_z) = (-y, x, 0) }

Construct a curve according to the following directions:

1) Start with a unit-circle in the x y xy plane with its center on the origin
2) Keep points ( 0 , 1 ) (0,-1) and ( 0 , 1 ) (0,1) pinned to the x y xy plane so they can't move
3) Bend the right ( x > 0 ) (x>0) half of the circle toward the + z +z direction so that it makes an angle of π 4 \frac{\pi}{4} with the x y xy plane. The points on the right side of the circle should still be co-planar after the bend.
4) Bend the left ( x < 0 ) (x<0) half of the circle toward the + z +z direction so that it makes an angle of π 4 \frac{\pi}{4} with the x y xy plane. The points on the left side of the circle should still be co-planar after the bend.

The two bending operations should yield a closed and continuous (but not continuously differentiable) curve. Determine the absolute value of the circulation of F \vec{F} over the curve.


The answer is 4.44288.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Otto Bretscher
Dec 15, 2018

Parameterise the bent curve as x = cos ( π 4 ) cos t x=\cos(\frac{\pi}{4})\cos t and y = sin t y=\sin t , for 0 t 2 π 0 \leq t \leq 2\pi (we will not need z z ). Now C y d x + x d y = 0 2 π 1 2 d t = 2 π 4.443 \int_C-ydx+xdy=\int_{0}^{2\pi}\frac{1}{\sqrt{2}}dt=\sqrt{2}\pi\approx \boxed{4.443} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...