Verified answer

Take spherical coordinates.- note that the radius is normal to the surface , so a unit normal is <n>= r/R

where r= RsinZcosT i+ RsinZsinT j+RcosZ k .- Note that we use R because we are moving on the surface of the sphere .-

so <n> = r/R =sinZcosT i+ sinZsinT j+cosZ k

Note also, that we are integrating over the surface directly ( No projection over any cartesian plane) , so the limits will be

0<Z<pi/2

0<T<2pi

Curl F = (0-0, 0-1,2-(-1) ) = (0,-1,3)

Curl F.dS is really curl F.<n> dS , where dS = RsinZdT (RdZ) = R^2sinZdzdT

CurlF.<n>dS= ( (0)*sinZcosT + (-1)*sinZsinT +(3)*cosZ )R^2sinZdZdT

CurlF.<n>dS = R^2 ( -sin^2Z sinT +3sinZcosZ) dZdT

0<Z<pi/2

0<T<2pi

By the Stoke´s Theorem this represents the work ( dW ) done by F along the curve x^2+y^2=1

z=1−x^2−y^2 is G(x,y,z) = z+x^2+y^2=1 and a normal to a surface dS will Nabla G

N= 2x i+2y j+k

Since we will project the normal to the surface over the plane XY , Region x^2+y^2=1 ,

<n>= N/INI and curl F.<n>dS= =curl F.N/INI dS .- This is the projection of curl F normal to the surface dS , now , the projection of dS over the palne XY will be <n>dS. k = dA , so dS = dA/ I<n>.k I

dS= dA/ I( N/INI .kI = INI dA /IN.kI , but N.k = 1 , so

dS= INIdA , thus , finally

curl F.<n>dS = curl F. (N/INI dS) =curl F. (N/INI) INI dA = curlF . NdA ( dA over XY plane)

curlF= (1-0,0-0,1-2x) = ( 1,0,1-2x)

CurlF .NdA=( 1,0,1-2x) . (2x i+2y j+k ) dA

curlF.NdA= (2x+0+1-2x) dA

curlF.dS= curlF.NdA = dA

Thus, INT_S curlF.dS= INT_R curlF.NdA = INT_R dA = pi (1)^2 = pi ( Over the Region , a circle with radius R=1)

By the Stoke´s Theorem this represents the work ( Work ) done by F along the curve x^2+y^2=1 too.-

S = unit sphere

F(x,y,z) = <-y, 2x, x+z>

I forgot how it works but this looks extremely familiar and correct to me, better than I would have remembered it: https://en.wikipedia.org/wiki/Curl_%28mathematics%...

including especially the way it is calculated

The normal vector to the sphere is 2xi + 2yj + 2zk

It should be a unit vector so we do that divided by 2√3

1/√3 (xi + yj + zk) and here i,j,k are meant to be the unit vectors on the axes to be a basis etc

make this matrix and do the determinant pattern

i...j...k

-y 2x x+z

x..y...z

i[(2xz - (x+z)y)] + j[(x+z)(x) - (-y)(z)] + k[(-y)y - (2x)x]

all divided by √3 don't forget

Then that has to be integrated over the surface.

I sort of think this theorem can apply and then it would come out to 0 because the surface doesn't have a border. The Kelvin-Stokes theorem. But also I am not sure if that is right, it has been a looooong time and it doesn't look the same as it used to.

https://en.wikipedia.org/wiki/Stokes%27_theorem#Ke...