Suppose F = <-y ,x ,z> and S is the part of sphere x^2 + y^2 + z^2 = 25 below the plane z=4, oriented with the outward-pointing normal

Question

(The normal at (5,0,0) is i). Compute the ∫∫ curl F . dS by using Stoke's Theorem. I already found the answer by using line integral but I want to know the answer by using the surface integral. Answer= -18pi

Answer 1

Surface:

 

Break it into the parts above the xy-plane and below it.

 

Above: Domain is the washer 9 ≤ x² + y² ≤ 25. Use two integrals here.

 

Vector: S=⟨ √(25 - x² - y²) / x , √(25 - x² - y²) / y, 1 ⟩

 

Curl F =

 

| i    j   k |

|D_x D_y D_z |

|-y  x   z  |

 

= ⟨0,0,2⟩

 

Curl F · dS = 2.

 

Upper part: ∫∫_U Curl F · dS = 2(25π - 9π) = 32π

 

Similarly, for the bottom part, but the signs of the dS vector components flip.

 

Curl F · dS = -2.

 

Lower part: ∫∫_L Curl F · dS = -2(25π) = -50π

 

So the total surface integral over the entire surface is 32π+(-50π) = -18π