Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
202 LECTURE 25. STOKES’ THEOREM Stokes’ theorem as we have presented here is really a special case in 3 dimensions. A more general form would have S ⊂ R n and ∂S ⊂ R n−1 , and would be written using ”generalized” surface and volume integrals as ∫ ∫ ω = dω (25.6) ∂S where dω represents the “exterior derivative” of a generalized vector field ω called a “differential form.” We have already met several special cases to equation 25.6: 1. In one dimension f : (a, b) ⊂ R ↦→ R, S is a line segment and ∂S is the set of endpoints {a, b}. The derivative is f ′ , so that ∫ b which is the fundamental theorem of calculus. a S f ′ (t)dt = f(b) − f(a) (25.7) 2. In two dimensions we can write the vector field as F = (M, N, 0), S is an area A ⊂ R 2 , ∂S is the boundary C of A, and n = (0, 0, 1) so that (∇ × F)) · n = N x − M y (25.8) Equation 25.5 then becomes ∮ ∮ F · dr = (Mdx + Ndy) = (N x − M y )dA (25.9) which is Green’s theorem. C C 3. In three dimensions S = V , ∂S = ∂V , ω = F · n and dω = ∇ · F, giving F · n dS = ∇ · F dV (25.10) ∂V which is the divergence theorem (also called Gauss’ theorem). Example 25.1 Find ∮ C F · dr for F = ( yz 2 − y, xz 2 + x, 2xyz ) where C is a cirlce of radius 3 centered at the origin in the xy-plane, using Stoke’s theorem. Solution. Letting S denote the disk that is enclosed by the circle C will can the formula ∮ F · T ds = (∇ × F) · n dS C S Revised December 6, 2006. Math 250, Fall 2006 A
LECTURE 25. STOKES’ THEOREM 203 Since C, and hence S, both lie in the xy-plane we have (n) = (0, 0, 1). Hence (∇ × F) · n = (P y − N z , M z − P x , N x − M y ) · (0, 0, 1) = N x − M y = ∂ ∂x (xz2 + x) − ∂ ∂y (yz2 − y) = z 2 + 1 − z 2 + 1 = 2 Hence ∮ C F · T ds = 2 S dS = 2π(3 2 ) = 18π Example 25.2 Find ∮ C F · dr for where C is a cirlce defined by using Stoke’s theorem. F = (z − 2y, 3x − 4y, z + 3y x 2 + y 2 = 4, z = 1 Solution We are again going to use the formula ∮ F · T ds = (∇ × F) · n dS C We again have n = (0, 0, 1), and C is a circle of radius 2. Hence S (∇ × F) · n = (P y − N z , M z − P x , N x − M y ) · (0, 0, 1) = N x − M y = ∂ ∂x (3x − 4y) − ∂ (z − 2y) ∂y 3 + 2 = 5 Hence ∮ C F · T ds = 5 S dS = 5π(1 2 ) = 20π Math 250, Fall 2006 Revised December 6, 2006.
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202 LECTURE 25. STOKES’ THEOREM<br />
Stokes’ theorem as we have presented here is really a special case in 3 dimensions.<br />
A more general form would have S ⊂ R n and ∂S ⊂ R n−1 , and would be written<br />
using ”generalized” surface and volume integrals as<br />
∫ ∫<br />
ω = dω (25.6)<br />
∂S<br />
where dω represents the “exterior derivative” of a generalized vector field ω called<br />
a “differential form.” We have already met several special cases to equation 25.6:<br />
1. In one dimension f : (a, b) ⊂ R ↦→ R, S is a line segment and ∂S is the set<br />
of endpoints {a, b}. The derivative is f ′ , so that<br />
∫ b<br />
which is the fundamental theorem of calculus.<br />
a<br />
S<br />
f ′ (t)dt = f(b) − f(a) (25.7)<br />
2. In two dimensions we can write the vector field as F = (M, N, 0), S is an<br />
area A ⊂ R 2 , ∂S is the boundary C of A, and n = (0, 0, 1) so that<br />
(∇ × F)) · n = N x − M y (25.8)<br />
Equation 25.5 then becomes<br />
∮ ∮<br />
<br />
F · dr = (Mdx + Ndy) = (N x − M y )dA (25.9)<br />
which is Green’s theorem.<br />
C<br />
C<br />
3. In three dimensions S = V , ∂S = ∂V , ω = F · n and dω = ∇ · F, giving<br />
<br />
<br />
F · n dS = ∇ · F dV (25.10)<br />
∂V<br />
which is the divergence theorem (also called Gauss’ theorem).<br />
Example 25.1 Find ∮ C<br />
F · dr for<br />
F = ( yz 2 − y, xz 2 + x, 2xyz )<br />
where C is a cirlce of radius 3 centered at the origin in the xy-plane, using Stoke’s<br />
theorem.<br />
Solution. Letting S denote the disk that is enclosed by the circle C will can the<br />
formula<br />
∮<br />
<br />
F · T ds = (∇ × F) · n dS<br />
C<br />
S<br />
Revised December 6, 2006. Math 250, Fall 2006<br />
A