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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    This is a problem with work and integration, which I'm not great at

  2. moazzam07
    • one year ago
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    https://answers.yahoo.com/question/index?qid=20100504211501AAq7ULx check this out

  3. Jhannybean
    • one year ago
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    |dw:1439775624180:dw|

  4. Jhannybean
    • one year ago
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    |dw:1439776368317:dw|

  5. Jhannybean
    • one year ago
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    Sorry, I was trying to work out an explanation.

  6. Jhannybean
    • one year ago
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    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..!!

  7. Jhannybean
    • one year ago
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    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 \(dy\)

  8. anonymous
    • one year ago
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    Okay, I'm following so far

  9. anonymous
    • one year ago
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    Do I not account for the trough's shape being y = x^8?

  10. anonymous
    • one year ago
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    Ohh okay

  11. anonymous
    • one year ago
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    So should it be \[W = \int\limits_{0}^{1} (62)*(1-y)*(6\sqrt[8]{y}) dy\]?

  12. Jhannybean
    • one year ago
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    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: \(2\sqrt[8]{y} \cdot 9 = 18\sqrt[8]{y}\) \[W=\int_0^1 (62~ft\cdot lbs) \cdot (1-y) \cdot (18\sqrt[8]{y}) \cdot dy\]

  13. anonymous
    • one year ago
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    Thanks Jhannybean!

  14. Jhannybean
    • one year ago
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    Do you see how ti works now?

  15. anonymous
    • one year ago
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    I do. It's still a tough concept for me though. Would you mind looking at a similar question for me?

  16. Jhannybean
    • one year ago
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    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 :)

  17. anonymous
    • one year ago
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    Thanks, you too!

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