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Note on September 5, 2006
Note on September 5, 2006
Problem #40 in Section 7.3 is interesting. It is a typical calculus problem in applications. Let
us see how can we solve the problem using our learned knowledge. We only need to consider the
following cross-section areas (why?):
5
2
0
2
5
5
2
0
2
5
x
y
crosssection of the water storage tank
Figure 7.3-HW40. Circular cross-section of the water storage tank. The boundary of the cross-section area
can be modeled via functions y = �
25 x
2
with the center (0, 0). The green line indicates the water level.
Needless to say that the total area of the above cross-section is
A = r
2
= 25.
(1.1)
As for the area of the cross-section under the water, according to the above gure, it would be
easier to calculate by using an integral in the Y direction (why?). Let this part of the area be B.
We have (why?)
B = A
2 + 2
2
0
25 y
2
dy.
(1.2)
Our purpose is to compute the percentage value of the ratio B/A. For solving (1.2), we let y =
5 cos u. Therefore, as y = 0, u = cos
1
(0) = /2; as y = 2, u = cos
1
(2/5). Further, dy =
5 sin udu. Substitute them into (1.2), according Example 3 on page 483, we acquire that
B =
25
2 2
y=2
y=0
25 25 cos
2
u5 sin udu
=
25
2 50
cos
1
(2/5)
/2
1 cos
2
u sin udu
=
25
2 + 50
/2
cos
1
(2/5)
sin
2
udu = 25
2 + 25 u
1
2 sin(2u)
/2
cos
1
(2/5)
=
25
2 + 25 2
1
2 sin() cos
1
2
5 +
1
2 sin 2 cos
1
2
5
58.7230.
Thus,
B
A
58.7230
25
0.7477 = 74.77%.
MTH 1322/Q. Sheng