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CHAP. 10]
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LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
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If U x; y; z 3z, (2) becomes q q 3z 1 4x2 4y2 dx dy 3f2 x2 y2 g 1 4x2 4y2 dx dy
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or in polar coordinates, 2 p2
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p 111 3 2 2 1 42 d d 10  0
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Physically this could represent the mass of S assuming a density 3z, or three times the rst moment of S about the xy plane.
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10.18. Find the surface area of a hemisphere of radius a cut o by a cylinder having this radius as diameter.
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Equations for the hemisphere and cylinder (see Fig. 10-15) 2 2 2 2 are given p respectively by x y z 2a 2 (or z a2 x2 y2 and x a=2 2 y2 a2 =4 (or x y ax). Since x zx p 2 a x2 y2 we have and y zy p 2 a x2 y2 Fig. 10-15
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q a p dx dy Required surface area 2 1 z2 z2 dx dy 2 x y a2 x2 y2
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Two methods of evaluation are possible. Method 1: Using polar coordinates. Since x2 y2 ax in polar coordinates is  a cos , the integral becomes =2 a cos  2
 0  0
a p  d d 2a a2 2
=2
p a cos   a2 2  d 
2a2 Method 2: The integral is equal to p 2
a ax x
=2
1 sin  d  2 a2
x 0 y 0
a p dy dx 2a a2 x2 y2
pax x2  y sin 1 p  dx  ax x2 y 0 x 0 r a x dx 2a sin1 a x 0 a
Letting x a tan2 , this integral becomes & ' =4 =4 4a2  tan  sec2 d 4a2 1  tan2 j=4 1 tan2  d 0 2 2
& ' =4 2a  tan2 j=4 sec2  1 d 0 0 n o 2a2 =4 tan   j=4  2 a2 0
Note that the above integrals are actually improper and should be treated by appropriate limiting procedures (see Problem 5.74, 5, and also 12).
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
[CHAP. 10
10.19. Find the centroid of the surface in Problem 10.17.
z dS " " By symmetry, x y 0 and
S " z
q z 1 4x2 4y2 dx dy
r q dS 1 4x2 4y2 dx dy r
The numerator and denominator can be obtained from the results of Problems 10.17(c) and 10.17(a), 37=10 111 " . respectively, and we thus have z 13=3 130
10.20. Evaluate
A n dS, where A xyi x2 j x z k, S is that portion of the plane
2x 2y z 6 included in the rst octant, and n is a unit normal to S. (See Fig. 10-16.)
A normal to S is r 2x 2y z 6 2i 2i 2j k 2i 2j k . Then 2j k, and so n p 2 22 12 3 2   2i 2j k A n fxyi x2 j x z kg 3 2xy 2x2 x z 3 2xy 2x2 x 6 2x 2y 3 2xy 2x2 x 2y 6 3
The required surface integral is therefore Fig. 10-16 ! ! q 2 2 2xy 2x x 2y 6 2xy 2x x 2y 6 dS 1 z2 z2 dx dy x y 3 3 r ! 2xy 2x2 x 2y 6 p 12 22 22 dx dy 3 3
3 x
2xy 2x2 x 2y 6 dy dx
x 0 3 x 0
xy2 2x2 y xy y2 6y j3 x dx 27=4 0
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 gure. If n is the positive normal at point P of the surface, we nd 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 nd the whole thing colored. This surface, called a Mo bius strip,
Fig. 10-17
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
is an example of a one-sided surface. This is sometimes called a nonorientable surface. A two-sided surface is orientable.
THE DIVERGENCE THEOREM 10.22. Prove the divergence theorem. (See Fig. 10-18.)
Fig. 10-18
Let S be a closed surface which is such that any line parallel to the coordinate axes cuts S in at most two points. Assume the equations of the lower and upper portions, S1 and S2 , to be z f1 x; y and z f2 x; y , respectively. Denote the projection of the surface on the xy plane by r. Consider
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