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Given the field [tex]H=\frac{1}{2}cos\frac{\phi}{2}\hat{\rho}-sin\frac{\phi}{2}\hat{\phi}[/tex],evaluate both sides of Stokes’ theorem for the path formed by the intersection of the cylinder [tex]\rho = 3[/tex] and the plane z = 2, and for the surface defined by [tex]\rho = 3[/tex], 0<z<2 , and z = 0, 0<[tex]\rho[/tex]<3.

I have problem at [tex]\frac{1}{2}cos\frac{\phi}{2}\hat{\rho}[/tex] part.

when i do line integral , the dot product of rho and phi will gives me zero

but when i do surface integral, after doing the curl, i will get z vector, dot with the surface with z vector also, i will obtain a value but not zero.

i cannot get LHS and RHS equal

Anyone can help me solve this, thanks!

I have problem at [tex]\frac{1}{2}cos\frac{\phi}{2}\hat{\rho}[/tex] part.

when i do line integral , the dot product of rho and phi will gives me zero

but when i do surface integral, after doing the curl, i will get z vector, dot with the surface with z vector also, i will obtain a value but not zero.

i cannot get LHS and RHS equal

Anyone can help me solve this, thanks!

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