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Math Help - work with integral

  1. #1
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    work with integral

    i m stuck on both of these please help..

    a.

    If a force of 90 N stretches a spring 1 m beyond its natural length. how much work does it take to stretch the spring 5 m beyond its natural length?



    b.

    to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??
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  2. #2
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    ANSWERS

    a.
    1125 joules

    b. about 21,446,605 j



    i am at a utter lost for words.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    i m stuck on both of these please help..

    a.

    If a force of 90 N stretches a spring 1 m beyond its natural length. how much work does it take to stretch the spring 5 m beyond its natural length?

    This is an application of Hookes law that says F=kx.

    We're initially told that it takes a force of 90 N to stretch it 1 m. This gives us enough info to find k, the spring constant:

    90=k(1)\implies k=90

    Thus, F=90x.

    Therefore, the total amount of work done to stretch it to 5 m is W=90\int_0^5 x\,dx

    Can you take it from here?

    b.

    to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??
    I don't have much time to do this one. I will come back later, if no one else has answered this.

    --Chris
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  4. #4
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    hm still need help on b, really strugglin.
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  5. #5
    MHF Contributor
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    Demonstrate your efforts, please. What have you tried? What did the integral look like? Why do you think it is incorrect? That is an awesome problem! You can learn a ton of calculus by solving it. Go!!
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  6. #6
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    Quote Originally Posted by TKHunny View Post
    Demonstrate your efforts, please. What have you tried? What did the integral look like? Why do you think it is incorrect? That is an awesome problem! You can learn a ton of calculus by solving it. Go!!
    here is what i did for b.. the right answer is 21 mil, i got 17mil


    to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??




    anyone have a clue??
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  7. #7
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    Take a good hard look at "16-x" vs your limits "[0,4]". You've pumped from 12 to 16. What about the rest?

    Personally, I would try this:

    \int_{0}^{16}\left[\pi x^{2}\right]\cdot (16 - y)\;dy\;=\;\int_{0}^{16}[\pi y] \cdot (16 - y)\;dy
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