Question

Let C be the straight line curve between the points (0, 2) and (1,0). Let N be the unit normal vector field on C, oriented so that it points away from the origin. Let F:R² → R² be the vector field defined by F(x, y) = (x, 17). Find the flux of F across the curve C in the direction pointing away from the origin.

Answer 1

The flux of the vector field across the curve in the direction pointing away from the origin is -35.

Step-by-step explanation:

To find the flux of the vector field F across the curve C in the direction pointing away from the origin, we can use the formula:

Flux = ∫C F · N ds

where F is the vector field, N is the unit normal vector field on C, and ds is the infinitesimal arc length along C.

In this case, F(x, y) = (x, 17), and N is oriented so that it points away from the origin. The curve C is the straight line segment between the points (0, 2) and (1, 0).

To calculate the integral, we need to parameterize the curve C and express ds in terms of the parameter.

We can parameterize the curve C as r(t) = (t, 2 - 2t) where 0 ≤ t ≤ 1.

Now we can calculate the flux:

Flux = ∫C F · N ds

= ∫01 (t, 17) · (-2, -2) dt

= ∫01 -2t + (-2)(17) dt

= ∫01 -2t - 34 dt

= -t2 - 34t

Evaluated from 0 to 1, the flux is:

Flux = -12 - 34(1) - (-02 - 34(0))

= -1 - 34

= -35