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Math Help - Force due to fluid pressure

  1. #1
    Member VitaX's Avatar
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    Force due to fluid pressure

    A window on the side of a 5 foot deep swimming pool is in the shape of an inverted parabola y=\frac{(9-x^2)}{3} located at the bottom of the pool 2 feet below the surface. Find the force F of water on this window.

    This formula is used: F=\int_a^b whA dy where w is the density of the fluid, h is the strip depth and A is L(y)dy.

    x =\sqrt{9 - 3y}

    F=\int_0^3 62.4(5-y)(2\sqrt{9-3y}) dy

    F= 124.8\left[\int_0^3 5\sqrt{9-3y} - \int_0^3 y\sqrt{9-3y}\right] dy

    I'm not sure if its supposed to be 2\sqrt{9-3y} or just \sqrt{9-3y} if someone can answer this part than I should be able to do the rest. However my friend says the first is u substitution but the 2nd one is integration by parts. What exactly is integration by parts?
    Last edited by VitaX; January 27th 2010 at 09:31 PM.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VitaX View Post
    so am I supposed to change the form of the given equation into terms of y instead of x?
    Yes
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Yes
    How is my work thus far in the problem?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VitaX View Post
    How is my work thus far in the problem?
    Yes, it should be 2\sqrt{9-x^2}, but where are you getting h=6-y?
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  5. #5
    Member VitaX's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Yes, it should be 2\sqrt{9-x^2}, but where are you getting h=6-y?
    Well in all my notes when the object is submerged in water and total height was taken to be from the height of the surface in this case 6 subtracting the height of an unknown strip length in the object which is y.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    But, this is what your post says:

    Quote Originally Posted by VitaX View Post
    A window on the side of a 5 foot deep swimming pool...
    I was just clarifying. You've got the idea. If it's 6, then use 6.
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  7. #7
    Member VitaX's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    But, this is what your post says:



    I was just clarifying. You've got the idea. If it's 6, then use 6.
    oh crap you are correct. I dont know where I got 6 from. Now that I've set the integral up correctly. Could you help me integrate this, I'm a little lost on the 2nd part, first part is clearly u sub right?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VitaX View Post
    oh crap you are correct. I dont know where I got 6 from. Now that I've set the integral up correctly. Could you help me integrate this, I'm a little lost on the 2nd part, first part is clearly u sub right?
    u=\sqrt{9-3y} imples that y=\frac{9-u^2}{3} so that dy=-\frac2{3}udu

    Can you see what to do?
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  9. #9
    Member VitaX's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    u=\sqrt{9-3y} imples that y=\frac{9-u^2}{3} so that dy=-\frac2{3}udu

    Can you see what to do?
    Hmmmm I see that you let u=\sqrt{9-3y} so that then you can say u=x and therefore sub u for x in the original function. But I dont really see how you found what dy is equal to and say I did understand this part, what would the next step be in the integral?
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VitaX View Post
    Hmmmm I see that you let u=\sqrt{9-3y} so that then you can say u=x and therefore sub u for x in the original function. But I dont really see how you found what dy is equal to and say I did understand this part, what would the next step be in the integral?
    Well, one thing at a time. I want you to understand what I've done.

    Let u=\sqrt{9-3y}

    Square both sides

    u^2=9-3y

    Solve for y

    y=\frac{9-u^2}{3}

    differentiate


    dy=-\frac{2}{3}udu

    Now substitute

    \int_0^3y\sqrt{9-3y}dy=\int_{y=0}^{y=3}\overbrace{\frac{9-u^2}{3}}^y\overbrace{(u)}^{\sqrt{9-3y}}\overbrace{\left(\frac{-2}{3}udu\right)}^{dy}
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  11. #11
    Member VitaX's Avatar
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    F = 124.8\left[\int_3^0 5u\left(\frac{-2}{3}u\right)\right - \int_3^0 \frac{9-u^2}{3}u\left(\frac{-2}{3}u\right)\right] du

    Is this correct so far?

    Crap accidently did a double post when editing an error
    Last edited by VitaX; January 27th 2010 at 10:55 PM.
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  12. #12
    Member VitaX's Avatar
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    F = 124.8\left[\int_3^0 5u  - \int_3^0 \frac{9-u^2}{3}(u)\left(\frac{-2}{3}u\right)\right] du

    Is this correct so far? Or is the \frac{-2}{3}u supposed to be in the first integral as well
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