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Math Help - 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 \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 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

    \int_0^h\pi a^2 dh
    Last edited by RockHard; December 5th 2009 at 10:48 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
    <br />
r(h) = a - | h - a | <br />
    <br />
r(0) = 0<br />
    <br />
r(a) = a<br />
    <br />
r(2a) = 0<br />
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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
    <br />
r(h) = a - | h - a | <br />
    <br />
r(0) = 0<br />
    <br />
r(a) = a<br />
    <br />
r(2a) = 0<br />
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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)
    <br />
r(h) = \sqrt(a^2-(a-h)^2) = \sqrt(2ah-h^2)<br />

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

    and the integral is not zero,

    you need to calculate

    <br />
\int_0^h\pi(2ak-k^2) dk = \pi \* (a\*k^2-k^3/3)|_0^h <br />
= \pi \* a\*h^2 - \pi \* h^3/3<br />
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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)
    <br />
r(h) = \sqrt(a^2-(a-h)^2) = \sqrt(2ah-h^2)<br />

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

    and the integral is not zero,

    you need to calculate

    <br />
\int_0^h\pi(2ak-k^2) dk = \pi \* (a\*k^2-k^3/3)|_0^h <br />
= \pi \* a\*h^2 - \pi \* h^3/3<br />
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