Page 257 - Schaum's Outline of Theory and Problems of Advanced Calculus

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248 LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS [CHAP. 10

10.19. Find the centroid of the surface in Problem 10.17.
ðð
ðð q
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2
2
zdS z 1 þ 4x þ 4y dx dy
S r
By symmetry, x ¼ y ¼ 0 and z z ¼ ðð ¼ ðð q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
x
y
2
2
dS 1 þ 4x þ 4y dx dy
S r
The numerator and denominator can be obtained from the results of Problems 10.17(c) and 10.17(a),
37 =10 111
z .
13 =3 130
respectively, and we thus have z ¼ ¼
ðð
2
10.20. Evaluate A n dS, where A ¼ xyi x j þðx þ zÞk, S is that portion of the plane
S
2x þ 2y þ z ¼ 6 included in the ﬁrst octant, and
n is a unit normal to S. (See Fig. 10-16.)
A normal to S is rð2x þ 2y þ z 6Þ¼ 2i þ
2i þ 2j þ k 2i þ 2j þ k
. Then
2j þ k, and so n ¼ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼
2
2
2 þ 2 þ 1 2 3

2 2i þ 2j þ k
3
A n ¼fxyi x j þðx þ zÞkg
2
2xy 2x þðx þ zÞ
3
¼
2
2xy 2x þðx þ 6 2x 2yÞ
¼
3
2
2xy 2x x 2y þ 6
¼
3
The required surface integral is therefore Fig. 10-16
! !
ðð 2 ðð 2 q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2xy 2x x 2y þ 6 2xy 2x x 2y þ 6
2
2
1 þ z x þ z y dx dy
3 dS ¼ 3
S r
ðð 2 !
2xy 2x x 2y þ 6 p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2
2
2
1 þ 2 þ 2 dx dy
3
¼
r
3 3 x
ð ð
2
ð2xy 2x x 2y þ 6Þ dy dx
¼
x¼0 y¼0
3
ð
2 2 2 3 x
ðxy 2x y xy y þ 6yÞj 0 dx ¼ 27=4
¼
x¼0
10.21. In dealing with surface integrals we have restricted
ourselves to surfaces which are two-sided. Give an
example of a surface which is not two-sided.
Take a strip of paper such as ABCD as shown in the
adjoining Fig. 10-17. Twist the strip so that points A and
B fall on D and C, respectively, as in the adjoining ﬁgure.
If n is the positive normal at point P of the surface, we
ﬁnd that as n moves around the surface, it reverses its
original direction when it reaches P again. If we tried
to color only one side of the surface, we would ﬁnd the
whole thing colored. This surface, called a Mo ¨bius strip, Fig. 10-17

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