Annly theorem o evaluate s to ke dr 2 dy x d 2 D cu there C

Annly theorem o evaluate s to ke dr 2 dy x d 2 D cu there C s e c curve of intersection of z -t -t 2

Solution

As per the given question, vector field is

F = y i + z j + x k.

Curl of F is:

curl(F) = -i - j - k.

Any surface that has this curve as it\'s boundary can be used with Stokes\' theorem.
We can use the plane x + z = a as the surface.

Denoting this as g(x, y, z) = a, the normal is the gradient.

g = i + k.

The unit normal (oriented upward) would then be

n = g/||g|| = (i + k)/(2).

The differential surface area

dS = [1 + (z/x)² + (z/y)²] dA = (2) dA

because z = a - x

z/x = -1

z/y = 0. This assumes that dA is a differential area element in the xy-plane such as dxdy or dydx.

So the integrand will be

curl(F)n dS = <-1, -1, -1><1, 0, 1> dA = -2 dA.

Using z = a - x and sub this into the sphere equation

a² = x² + y² + (a - x)² = 2x² - 2ax + y² + a² ==> 2x² - 2ax + y² = 0.

So the projection of the intersection into the xy-plane gives an ellipse

2(x - a/2)² + y² = a²/2 ==> (x - a/2)²/(a/2)² + y²/(a/(2))² = 1.

This has the form (x - h)²/A² + y²/B² = 1 which is an ellipse centered at (h, 0) with area

Area = AB.
A = a/2 and B = a/(2), for an area of Area = a²/[2(2)].

So finally, the integral is

curl(F)n dS = - a²/(2).