Page 263 - Calculus Demystified

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SOLUTION CHAPTER 8 Applications of the Integral
As shown in Fig. 8.40, at depth h (measured down from the surface of the
water), the tank has width corresponding to the base of an isosceles triangle
similar to the triangle describing the end of the tank. The height of this triangle
is 5 − h. Thus we can solve
w(h) 4
= .
5 − h 9
We find that
4
w(h) = (5 − h).
9
According to (∗), the total force on the side is then

5 4
h · 62.4 · (5 − h) dh ≈ 577.778 lbs.
0 9

4

9
5
_
5 h
Fig. 8.40

EXAMPLE 8.28
An aquarium tank isfilled with a mixture of water and algicide to keep the
liquid clear for viewing.The liquid hasa density of 50 poundsper cubic foot.
For viewing purposes, a window is located in the side of the tank, with center
20 feet below the surface. The window is in the shape of a square of side
√
4 2 feet with vertical and horizontal diagonals(see Fig. 8.41). What isthe
total force on thiswindow?

SOLUTION
As usual, we measure depth downward from the surface with independent
variable h. Of course the square window has diagonal 4 feet. Then the range of
integration will be h = 20−4 = 16 to h = 20+4 = 24. Refer to Fig. 8.42. For
h between 16 and 20, we notice that the right triangle in Fig. 8.42 is isosceles
and hence has base of length h − 16. Therefore

w(h) = 2(h − 16) = 2h − 32.