Thomas Calculus 13th [Solutions]

476 Chapter 6 Applications of Definite Integrals
33. To find the width of the plate at a typical depth y, we first find an equation for the line of the plates right-hand
edge: y x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x 5 y and
the total width is L( y) 2x 2(5 y ). The depth of the strip is ( y ). The force exerted by the water against
2 2
one side of the plate is therefore F w( y) L( y) dy 62.4 ( y) 2(5 y)
dy
5 5
124.8 2 2 2 3
2
5 1 5 1 5 1
5 5 y y dy 124.8 y y
2 3 124.8 5
2 4 3 8 2 25 3
125
(124.8) 105 117 (124.8) 315 234 1684.8 lb
2 3 6
34. An equation for the line of the plates right-hand edge is y x 3 x y 3. Thus the total width is
L( y) 2x 2( y 3). The depth of the strip is (2 y ). The force exerted by the water is
0 0 0 2
F w(2 y) L( y) dy 62.4 (2 y) 2(3 y) dy 124.8 6 y y dy 124.8 6y
3 3 3
( 124.8) 18 9 9 ( 124.8) 27 1684.8 lb
2 2
35. (a) The width of the strip is L( y ) 4, the depth of the strip is
2 3 0
y y
2 3
b strip
(10 y) F w depth F( y)
dy
a
3 3
y
62.4(10 y)(4) dy 249.6 (10 y) dy 249.6 10 y 249.6 30 9 6364.8 lb
0 0
2 2
0
b strip
(b) The width of the strip is L( y ) 3, the depth of the strip is (10 y) F w depth F( y)
dy
a
4 4
y
62.4(10 y)(3) dy 187.2 (10 y) dy 187.2 10y
187.2(40 8) 5990.4 lb
0 0
2
0
36. The width of the strip is
2 3
2 4
2
L( y ) 2 25 y , the depth of the strip is
b strip
(6 y) F w depth F( y)
dy
a
5 2 5 2 5 2 5 2
62.4 (6 y) 2 25 y dy 124.8 (6 y) 25 y dy 124.8 6 25 y dy y 25 y dy
0 0 0 0
To evaluate the first integral, we use we can interpret
radius is 5, thus
5 2
25 y dy as the area of a quarter circle whose
0
5 2 5 2 1 2 75
0 6 25 y dy 6 0
25 y dy 6 4 (5) 2
. To evaluate the second integral let
2
5 2 0
u 25 y du 2 y dy;
y 0. u 25, y 5 u 0, thus y 25 y dy 1 u du
0 2 25
25 1/2 3/2
25
1 1
125
5 2 5 2
u du u . Thus, 124.8 6 25 y dy y 25 y dy 124.8 75 125
2 0 3 0 3
0 0
2 3
9502.7 lb.
37. Using the coordinate system of Exercise 32, we find the equation for the line of the plates right-hand edge to
y 4
be y 2x 4 x and L( y) 2x y 4. The depth of the strip is (1 y).
2
(a)
(b)
0 0 0 2
3y
y
F w(1 y) L( y) dy 62.4 (1 y)( y 4) dy 62.4 4 3y y dy 62.4 4y
4 4 4
2 3
F
(3)(16) 64 64 ( 62.4)( 120 64)
( 62.4) ( 4)(4) ( 62.4)( 16 24 ) 1164.8 lb
2 3 3 3
(3)(16) 64 ( 64.0)( 120 64)
( 64.0) ( 4)(4) 1194.7 lb
2 3 3
3
2 3 0
4
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