Calculus- Early Transcendentals, 2021a

546 Vector Calculus
Exercise 16.3.23 Find an f so that ∇ f = 〈x 3 ,−y 4 〉, or explain why there is no such f .
Exercise 16.3.24 Find an f so that ∇ f = 〈xe y ,ye x 〉, or explain why there is no such f .
Exercise 16.3.25 Find an f so that ∇ f = 〈ycosx,ysinx〉, or explain why there is no such f .
Exercise 16.3.26 Find an f so that ∇ f = 〈ycosx,sinx〉, or explain why there is no such f .
Exercise 16.3.27 Find an f so that ∇ f = 〈x 2 y 3 ,xy 4 〉, or explain why there is no such f .
Exercise 16.3.28 Find an f so that ∇ f = 〈yz,xz,xy〉, or explain why there is no such f .
∫
∫
Exercise 16.3.29 Evaluate (10x 4 −2xy 3 )dx− 3x 2 y 2 dy where C is the part of the curve x 5 −5x 2 y 2 −
C
C
7x 2 = 0 from (0,0) to (3,2).
Exercise 16.3.30 Let F = 〈yz,xz,xy〉. Find the work done by this force field on an object that moves from
(1,0,2) to (1,2,3).
Exercise 16.3.31 Let F = 〈e y ,xe y + sinz,ycosz〉. Find the work done by this force field on an object that
moves from (0,0,0) to (1,−1,3).
Exercise 16.3.32 Let
〈
〉
−x
F =
(x 2 + y 2 + z 2 ) 3/2 , −y
(x 2 + y 2 + z 2 ) 3/2 , −z
(x 2 + y 2 + z 2 ) 3/2
Find the work done by this force field on an object that moves from (1,1,1) to (4,5,6).
16.4 Green’s Theorem
We now come to the first of three important theorems that extend the Fundamental Theorem of Calculus
to higher dimensions. (The Fundamental Theorem of Line Integrals has already done this in one way, but
in that case we were still dealing with an essentially one-dimensional integral.) They all share with the
Fundamental Theorem the following rather vague description: To compute a certain sort of integral over
a region, we may do a computation on the boundary of the region that involves one less integration.
Note that this does indeed describe the Fundamental Theorem of Calculus and the Fundamental Theorem
of Line Integrals: to compute a single integral over an interval, we do a computation on the boundary
(the endpoints) that involves one less integration, namely, no integrations at all.
Theorem 16.12: Green’s Theorem
If the vector field f = 〈 f 1 , f 2 〉 and the region D are sufficiently nice, and if C is the boundary of D
(C is a closed curve), then
∫∫
D
∂ f 2
∂x − ∂ f 1
∂y
∫C
dA = f 1 dx+ f 2 dy,
provided the integration on the right is done counter-clockwise around C.