# MATH251Section16.8

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16.8 Stokes' Theorem Whereas Green's Theorem relates a double integral over a plane region D to a line integral around its plane boundary curve , Stokes' Theorem relates a surface integral over a surface S to a line integral around the boundary curve of S (which is a space curve). Let C be the boundary curve of an oriented surface S with the unit normal vector n . The positive orientation of C is defined as the counterclockwise direction centered at n . That is, the positive orientation of C is the direction of the four remaining fingers when the thumb is in the direction of n in the right hand rule . Stokes' Theorem Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation . Let F be a vector field whose components have continuous partial derivatives on an open region in R 3 that contains S . Then C F · d r = S curl F · d S . Since C F · d r = C F · T ds and S curl F · d S = S curl F · n dS, Stokes' Theorem says that the line integral around the boundary curve of S of the tangential com- ponent of F is equal to the surface integral over S of the normal component of the curl of F . The positively oriented boundary curve of the oriented surface S is often written as ∂S , so Stokes' Theorem can be expressed as S curl F · d S = ∂S F · d r as well. 1
EXAMPLE 1 Use Stokes' Theorem to evaluate C F · d r , where F ( x, y, z ) = - y 2 i + x j + z 2 k and C is the curve of intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1. (Orient C to be counterclockwise when viewed from above.) EXAMPLE 2 Use Stokes' Theorem to compute the integral S curl F · d S , where F ( x, y, z ) = xz i + yz j + xy k and S is the part of the sphere x 2 + y 2 + z 2 = 4 that lies inside the cylinder x 2 + y 2 = 1 and above the xy -plane. 2