Using stoke theorem evaluate integral F. dr given that F(x, y,z) = z^2i + 2xj + y^2k and the normal surface is given by s: z = 1-x^2-y^2

Your description of the surface is incomplete. But it looks like you're considering some subset of the paraboloid z = 1 - x ² - y ², so I'll go ahead and assume it's the part of said paraboloid above the x,y-plane, so that the boundary is a circle centered at the origin with radius 1.

By Stokes' theorem, the line integral of F along this boundary (∂S) is equal to the surface integral of curl(F ) over the surface itself (S). We have

F(x, y, z) = z ² i + 2x j + y ² k

which has curl

curl(F ) = (∂(y ²)/∂y - ∂(2x)/∂z) i - (∂(y ²)/∂x - ∂(z ²)/∂z) j + (∂(2x)/∂x - ∂(z ²)/∂y) k

curl(F ) = 2y i + 2z j + 2k

Parameterize S by the vector function,

r(u, v)  = u cos(v) i + u sin(v) j + (1 - u ²) k

with 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π.

Take the upward-pointing normal vector to S to be

n = ∂r/∂v × ∂r/∂u

n = (-u sin(v) i + u cos(v) j ) × (cos(v) i + sin(v) j - 2u k)

n = 2u ² cos(v) i + 2u ² sin(v) j + u k

Then the integral of curl(F ) over S - and hence the line integral of F over ∂S - is

Just to confirm this result, we can compute the line integral directly, since it's not so difficult to deal with. Parameterize ∂S by the vector function

r(t) = cos(t ) i + sin(t ) j

with 0 ≤ t ≤ 2π. (Note that there is a k component, but its coefficient is 0.) Then

dr/dt = -sin(t ) i + cos(t ) j

and the line integral is again

won't that be plagrisim thought?

step-by-step explanation:

nope.   happens to me as well.

-aparri