Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. Let S 1 be the portion of Ω(S) between S(a) and S,andlet∂S 1 be its boundary. Also let S L be the lateral surface of S 1 [that
r · n
r
is, the surface of S 1 except S and S(a)]. Applying the Divergence Theorem we have
dS = ∇ ·
∂S 1
r 3 S 1
r dV . 3
But
⇒
∇ ·
∂S 1
r · n
r 3 dS =
r ∂
r = 3 ∂x , ∂ ∂y , ∂
x
·
∂z (x 2 + y 2 + z 2 ) , y
3/2 (x 2 + y 2 + z 2 ) , z
3/2 (x 2 + y 2 + z 2 ) 3/2
= (x2 + y 2 + z 2 − 3x 2 )+(x 2 + y 2 + z 2 − 3y 2 )+(x 2 + y 2 + z 2 − 3z 2 )
(x 2 + y 2 + z 2 ) 5/2 =0
S 1
0 dV =0. On the other hand, notice that for the surfaces of ∂S 1 other than S(a) and S,
r · n =0 ⇒
r · n
0=
dS =
∂S 1
r 3
r · n
r 3
S
S
r · n
r · n
r · n r · n
r · n
dS +
dS +
dS = dS +
dS ⇒
r 3 S(a) r 3 S L
r 3 S r 3 S(a) r 3
r · n
dS = −
dS. NoticethatonS(a), r = a ⇒ n = − r
S(a) r 3 r = − r a and r · r = r2 = a 2 ,so
r · n
r · r
that −
dS =
S(a) r 3 S(a) a dS = a
S(a)
2
4 a dS = 1 area of S (a)
dS = = |Ω(S)|.
4 a
S(a)
2 a 2
r · n
Therefore |Ω(S)| = dS.
S r 3
3. The given line integral 1 (bz − cy) dx +(cx − az) dy +(ay − bx) dz canbeexpressedas F · dr if we define the vector
2 C C
field F by F(x, y, z) =P i + Q j + R k = 1 (bz − cy) i + 1 (cx − az) j + 1 (ay − bx) k. Thendefine S to be the planar
2 2 2
interior of C,soS is an oriented, smooth surface. Stokes’ Theorem says F · dr = curl F · dS = curl F · n dS.
C S S
Now
curl F =
∂R
∂y − ∂Q ∂P
i +
∂z ∂z − ∂R ∂Q
j +
∂x ∂x − ∂P
k
∂y
= 1
2 a + 1 2 a i + 1
2 b + 1 2 b j + 1
2 c + 1 2 c k = a i + b j + c k = n
so curl F · n = n · n = |n| 2 =1, hence curl F · n dS = dS whichissimplythesurfaceareaofS. Thus,
S S
C F · dr = 1 2
C (bz − cy) dx +(cx − az) dy +(ay − bx) dz is the plane area enclosed by C. 307