Calculus- Early Transcendentals, 2021a

566 Vector Calculus
∫ b
=
∫
=
a
∂E
[( ) ( ) ]
∂z dx
f 1 + f 3
∂x dt + ∂z dy
f 2 + f 3 dt
∂y dt
( ) ( )
∂z
∂z
f 1 + f 3 dx+ f 2 + f 3 dy,
∂x
∂y
which now looks just like the line integral of Green’s Theorem, except that the functions f 1 and f 2 of
Green’s Theorem have been replaced by the more complicated f 1 + f 3 (∂z/∂x) and f 2 + f 3 (∂z/∂y). We
can apply Green’s Theorem to get
∫
∂E
( )
∂z
f 1 + f 3
∂x
dx+
( )
∂z
f 2 + f 3
∂y
∫∫
dy =
E
( )
∂ ∂z
f 2 + f 3 − ∂ ( )
∂z
f 1 + f 3 dA.
∂x ∂y ∂y ∂x
Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes
∫∫
∂ f 2
E ∂x + ∂ f 2
∂z g x + ∂ f 3
∂x g y + ∂ f (
3
∂
∂z g f1
xg y + f 3 g yx −
∂y + ∂ f 1
∂z g y + ∂ f 3
∂y g x + ∂ f )
3
∂z g yg x + f 3 g xy dA
∫∫
∂ f 2
=
∂x + ∂ f 2
∂z g x + ∂ f 3
∂x g y − ∂ f 1
∂y − ∂ f 1
∂z g y − ∂ f 3
∂y g x dA,
E
which is the same as the expression we obtained for the surface integral.
♣
Exercises for 16.8
Exercise 16.8.1 Let f = 〈z,x,y〉. The plane z = 2x + 2y − 1 and the paraboloid z = x 2 + y 2 intersect in a
closed curve. Stokes’ Theorem implies that
∫∫
∮ ∫∫
(∇ × f) · NdS = f · dr = (∇ × f) · NdS
D 1 C
D 2
where the line integral is computed over the intersection C of the plane and the paraboloid, and the two
surface integrals are computed over the portions of the two surfaces that have boundary C (provided, of
course, that the orientations all match). Compute all three integrals.
Exercise 16.8.2 Let D be∫∫the portion of z = 1 − x 2 − y 2 above the xy-plane, oriented up, and let f =
〈xy 2 ,−x 2 y,xyz〉. Compute (∇ × f) · NdS.
D
Exercise 16.8.3 ∫ Let D be the portion of z = 2x + 5y insidex 2 + y 2 = 1, oriented up, and let f = 〈y,z,−x〉.
Compute f · dr.
∂D
x 2 zdx+3xdy−y 3 dz, where C is the unit circle x 2 +y 2 = 1 oriented counter-
∮
Exercise 16.8.4 Compute
clockwise.
C
Exercise 16.8.5 Let D be the portion of z = px + qy + r over a region in∫the xy-plane that has area A,
oriented up, and let f = 〈ax + by + cz,ax + by + cz,ax + by + cz〉. Compute f · dr.
∂D