anonymous
  • anonymous
A trough is 9 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
This is a problem with work and integration, which I'm not great at
moazzam07
  • moazzam07
https://answers.yahoo.com/question/index?qid=20100504211501AAq7ULx check this out
Jhannybean
  • Jhannybean
|dw:1439775624180:dw|

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Jhannybean
  • Jhannybean
|dw:1439776368317:dw|
Jhannybean
  • Jhannybean
Sorry, I was trying to work out an explanation.
Jhannybean
  • Jhannybean
So the trough is full of water, and we're calculating the amount of work it would take to empty out THE WATER from the trough...so..!!
Jhannybean
  • Jhannybean
If we make thin vertical slabs parallel to the ends of the trough, we will technically calculate the area of the trough. In calculating how much water needs to be removed, let's take a closer look at what a small section of the trough looks like with water. |dw:1439777280727:dw| Now we see a little slab of the trough cut vertically with water in it, and we also see in order to get the water out, we have to integrate with respect to y, therefore \(dy\)
anonymous
  • anonymous
Okay, I'm following so far
anonymous
  • anonymous
Do I not account for the trough's shape being y = x^8?
anonymous
  • anonymous
Ohh okay
anonymous
  • anonymous
So should it be \[W = \int\limits_{0}^{1} (62)*(1-y)*(6\sqrt[8]{y}) dy\]?
Jhannybean
  • Jhannybean
If you imagine the water inside the trough as a big rectangle, then finding the surface area of the water is easy. Because we're solving it in terms of y, we need to put everything in terms of y. \[y=x^8 \longrightarrow x=\sqrt[8]{y}\]\(x=\sqrt[8]{y}\) would only give us half the trough, so in order to represent the entire trough in terms of x, we add on another \(\sqrt[8]{y}\) width of trough: \(2\sqrt[8]{y}\) length of trough: \(9\) ft SA of trough: \(2\sqrt[8]{y} \cdot 9 = 18\sqrt[8]{y}\) \[W=\int_0^1 (62~ft\cdot lbs) \cdot (1-y) \cdot (18\sqrt[8]{y}) \cdot dy\]
anonymous
  • anonymous
Thanks Jhannybean!
Jhannybean
  • Jhannybean
Do you see how ti works now?
anonymous
  • anonymous
I do. It's still a tough concept for me though. Would you mind looking at a similar question for me?
Jhannybean
  • Jhannybean
I've actually got to head off now, got some college prep I need to do for tomorrow. Hopefully you'll get all your questions answered! Have a good day :)
anonymous
  • anonymous
Thanks, you too!

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