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Thread: volume of water in spherical tank funtion of depth double integral

  1. #1
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    Exclamation volume of water in spherical tank funtion of depth double integral

    Please if anyone could help I would appreciate it so much

    A water tank is in the shape of a sphere of radius a. Find the volume of the water as a function of the depth of the water h in the tank.

    I think that the best way to solve it would be to use double integrals.... I cannot figure it out however, any help at all I would appreciate.

    Thank you so much
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  2. #2
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    I don't believe you need to use a double integral, just partition the sphere on the interval 0 to the h(depth) of the tank, and to find the volume you calculate the integral of the area on the interval 0 to h I believe, can someone confirm this as I am not 100%

    Possible consider $\displaystyle \int_0^h\pi r^2 dh$

    I based this off if the tank was standing on the x-y plane, with y having relation to h and your partitions are being taken from 0 to some point h on the h-axis so to speak which can be related to some depth h, which the change in thickness of your partitions will be $\displaystyle dh$ Also note to determine your radius it needs to be in terms of h, instead of the usually given function that is in terms of x which remember I stated h has relation to y-axis, since your radius is a, I assume your integral would look like this

    $\displaystyle \int_0^h\pi a^2 dh$
    Last edited by RockHard; Dec 5th 2009 at 09:48 PM.
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  3. #3
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    Exclamation unsure

    Okay I think that if the depth is 0 the volume should be 0 and when the depth is 2a (the tank is full), the volume would be 4/3*pi*a^3. when i tried to use the integral that you suggested I might be doing something wrong but i cannot find get that answer. any help?
    Thanks so much for your answer i appreciate all the work that you have done to try and solve it i think that maybe you are right but I'm not sure if i am just doing something wrong!
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  4. #4
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    r is a function of h
    where
    $\displaystyle
    r(h) = a - | h - a |
    $
    $\displaystyle
    r(0) = 0
    $
    $\displaystyle
    r(a) = a
    $
    $\displaystyle
    r(2a) = 0
    $
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  5. #5
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    Exclamation

    I am confused. How did you get this? and if what you are saying is right then the integral above cannot be the correct integral becuase r(2a)=0, right?

    thank you so much!

    Quote Originally Posted by mooshazz View Post
    r is a function of h
    where
    $\displaystyle
    r(h) = a - | h - a |
    $
    $\displaystyle
    r(0) = 0
    $
    $\displaystyle
    r(a) = a
    $
    $\displaystyle
    r(2a) = 0
    $
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  6. #6
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    Quote Originally Posted by seipc View Post
    I am confused. How did you get this? and if what you are saying is right then the integral above cannot be the correct integral becuase r(2a)=0, right?

    thank you so much!
    apperently i was wrong about r(h)
    $\displaystyle
    r(h) = \sqrt(a^2-(a-h)^2) = \sqrt(2ah-h^2)
    $

    but the values on the edge i gave u are still the same

    and the integral is not zero,

    you need to calculate

    $\displaystyle
    \int_0^h\pi(2ak-k^2) dk = \pi \* (a\*k^2-k^3/3)|_0^h
    = \pi \* a\*h^2 - \pi \* h^3/3
    $
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  7. #7
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    Exclamation

    Okay i follow you all the way through this time, however, i am still confused on one thing. I know that the answer has to be 4/3*pi*a^3 when its at 2a (full) but i dont get that answer.... am i just doing something wrong.
    Thank you soooo much!

    Quote Originally Posted by mooshazz View Post
    apperently i was wrong about r(h)
    $\displaystyle
    r(h) = \sqrt(a^2-(a-h)^2) = \sqrt(2ah-h^2)
    $

    but the values on the edge i gave u are still the same

    and the integral is not zero,

    you need to calculate

    $\displaystyle
    \int_0^h\pi(2ak-k^2) dk = \pi \* (a\*k^2-k^3/3)|_0^h
    = \pi \* a\*h^2 - \pi \* h^3/3
    $
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