A trough is 7 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y = 8 from x = -1 to x = 1. The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot. - (-1)³)| 62 9

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I don't understand the work in the problem in the second image. (This is an answer from an expert). I substituted the 9 for 7. I just dont understand y^9/8 and y^17/8. I tried different ways to solve this problem to no avail. I'm at attempt 13.

A trough is 7 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is
shaped like the graph of y = 8 from x = -1 to x = 1. The trough is full of water. Find the
amount of work in foot-pounds required to empty the trough by pumping the water over the top.
Note: The weight of water is 62 pounds per cubic foot.
- (-1)³)|
62
9
Transcribed Image Text:A trough is 7 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y = 8 from x = -1 to x = 1. The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot. - (-1)³)| 62 9
The force is weight x volume
distance to which the volume element is to be lifted is 1-y.
Since we have
y = x8, - 1 ≤ x ≤ 1w = 62 pound/ ft.³
Now, the amount of work can be calculated by
The weight of the water. (volume of the elemnt) (distance to be lift
Now, transform the value of y in terms of x to obtain the the function in 'x'.
y = x³x = 8√y
Therefore, the work is
62.9.2ydy (1-y) = f 62·9·2·1 - ydy
The strip has to be lifted at (1-y) distance. We need to integrate the small work from y = 0 to y
= 1, we get
W =
=
=
||
W =
01116-1-ydy
1116 0.1-ydy
9
y =
1116-yldy
0
9
y y
111617
8
8 0
8
8
1116-17
466.82
Transcribed Image Text:The force is weight x volume distance to which the volume element is to be lifted is 1-y. Since we have y = x8, - 1 ≤ x ≤ 1w = 62 pound/ ft.³ Now, the amount of work can be calculated by The weight of the water. (volume of the elemnt) (distance to be lift Now, transform the value of y in terms of x to obtain the the function in 'x'. y = x³x = 8√y Therefore, the work is 62.9.2ydy (1-y) = f 62·9·2·1 - ydy The strip has to be lifted at (1-y) distance. We need to integrate the small work from y = 0 to y = 1, we get W = = = || W = 01116-1-ydy 1116 0.1-ydy 9 y = 1116-yldy 0 9 y y 111617 8 8 0 8 8 1116-17 466.82
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