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Chapter 14: Q. 2 (page 1134)
How might we generalize Green’s Theorem to higher dimensions? What sort of relationship might we hope to find between a triple integral over a (reasonably well behaved) three-dimensional region of space and a double integral over the surface that is the boundary of this region?
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Find the work done by the vector field
in moving an object around the unit circle, starting and ending at .
Find the masses of the lamina:
The lamina occupies the region of the hyperbolic saddle with equation that lies above and/or below the disk of radius 2 about the origin in the XY-plane where the density is uniform.
Compute dS for your parametrization in Exercise 9.
Work as an integral of force and distance: Find the work done in moving an object along the x-axis from the origin to if the force acting on the object at a given value of x is role="math" localid="1650297715748" .
Q. True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.
(a) True or False: Stokes’ Theorem asserts that the flux of a vector field through a smooth surface with a smooth boundary is equal to the line integral of this field about the boundary of the surface.
(b) True or False: Stokes’ Theorem can be interpreted as a generalization of Green’s Theorem.
(c) True or False: Stokes’ Theorem applies only to conservative vector fields.
(d) True or False: Stokes’ Theorem is always used as a way to evaluate difficult surface integrals.
(e) True or False: Stokes’ Theorem can be interpreted as a generalization of the Fundamental Theorem of Line Integrals.
(f) True or False: If F(x, y ,z) is a conservative vector field, then Stokes’ Theorem and Theorem 14.12 together give an alternative proof of the Fundamental Theorem of Line Integrals for simple closed curves.
(g) True or False: Stokes’ Theorem can be interpreted as a generalization of the Fundamental Theorem of Calculus.
(h) True or False: Stokes’ Theorem can be used to evaluate surface area .
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