Page 255 - Schaum's Outline of Theory and Problems of Advanced Calculus

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

q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2
F x þ F y þ F z
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2
j sec
j¼ 1 þ z x þ z y ¼
jF z j
according as the equation for S is z ¼ f ðx; yÞ or Fðx; y; zÞ¼ 0.
If the equation of S is Fðx; y; zÞ¼ 0, a normal to S at ðx; y; zÞ is rF ¼ F x i þ F y j þ F z k. Then
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
rF k ¼jrFjjkj cos
or F z ¼ F x þ F y þ F z cos
2
2
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2
F x þ F y þ F z
as required.
from which j sec
j¼
jF z j
In case the equation is z ¼ f ðx; yÞ,wecan write Fðx; y; zÞ¼ z f ðx; yÞ¼ 0, from which
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
1 þ z x þ z y .
F x ¼ z x ; F y z y ; F z ¼ 1 and we find j sec
j¼
ðð
2
2
10.17. Evaluate Uðx; y; zÞ dS where S is the surface of the paraboloid z ¼ 2 ðx þ y Þ above the xy
S
2
2
plane and Uðx; y; zÞ is equal to (a)1, (b) x þ y ,(c)3z. Give a physical interpretation in
each case. (See Fig. 10-14.)
The required integral is equal to
ðð
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
Uðx; y; zÞ 1 þ z x þ z y dx dy ð1Þ
r
where r is the projection of S on the xy plane given by
2
2
x þ y ¼ 2; z ¼ 0.
Since z x ¼ 2x; z y ¼ 2y,(1)can be written
ðð
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
Uðx; y; zÞ 1 þ 4x þ 4y dx dy ð2Þ
r
(a)If Uðx; y; zÞ¼ 1, (2)becomes
ðð q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fig. 10-14
2
2
1 þ 4x þ 4y dx dy
r
To evaluate this, transform to polar coordinates
ð ; Þ. Then the integral becomes
ffiffi
ffiffi p
p
ð 2 ð 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 2 1 2 13
2 2 3=2
1 þ 4 d d ¼ ð1 þ 4 Þ d ¼ 3
¼0 ¼0 ¼0 12 ¼0
Physically this could represent the surface area of S,orthe mass of S assuming unit density.
ðð
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
2
2
(b)If Uðx; y; zÞ¼ x þ y ,(2)becomes ðx þ y Þ 1 þ 4x þ 4y dx dy or in polar coordinates
r
ffiffi
p
ð 2 ð 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 149
2
3 1 þ 4 d d ¼
¼0 ¼0 30
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
where the integration with respect to is accomplished by the substitution 1 þ 4 ¼ u.
Physically this could represent the moment of inertia of S about the z-axis assuming unit density,
2
2
or the mass of S assuming a density ¼ x þ y .