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Thread: multiple integrals

  1. #1
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    multiple integrals

    Find the volume enclosed by the following regions:
    x^2 + y^2 = 1;
    y^2 + z^2 = 1;
    z^2 + x^2 = 1:
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  2. #2
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    Google "Steinmetz solid" for information and proofs.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Andreamet View Post
    Find the volume enclosed by the following regions:
    $\displaystyle {x^2} + {y^2} = 1$;
    $\displaystyle {y^2} + {z^2} = 1$;
    $\displaystyle {z^2} + {x^2} = 1$.
    You are given three circular cylinders, each of which is located along its coordinate axis. The axis of the first cylinder coincides with the axis $\displaystyle OZ$; second cylinder axis coincides with the axis $\displaystyle OX$; third cylinder axis coincides with the axis $\displaystyle OY$. We have a symmetry, so we should calculate the amount of body parts in the first oktant, and the answer multiplied by $\displaystyle 8$. Next, we consider the case $\displaystyle x \geqslant y$. Then the upper limit of integration for $\displaystyle z$ we can determine from the equation $\displaystyle {z^2} + {x^2} = 1$ which gives us a lower value for $\displaystyle z$, other than equation $\displaystyle {z^2} + {y^2} = 1$. The field of integration $\displaystyle D$ in the plane $\displaystyle XOY$, we can determine from the equation $\displaystyle {x^2} + {y^2} = 1$, i.e. this area is 1/8 of unit circle.

    So we have $\displaystyle V = 16\iint\limits_D {dxdy}\int\limits_0^{\sqrt {1 - {x^2}} } {dz} = 16\iint\limits_D {\sqrt {1 - {x^2}} dxdy}$, $\displaystyle D = \left\{ {\left. {x,y} \right|{\text{ }}{x^2} + {y^2} = 1,{\text{ }}0 \leqslant y \leqslant x} \right\}$.

    Next we will use cylindrical coordinates to calculate the integral

    $\displaystyle V = 16\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.
    \kern-\nulldelimiterspace} 4}} {d\phi } \int\limits_0^1 {\rho \sqrt {1 - {\rho ^2}{{\cos }^2}\phi } d\rho } = $

    $\displaystyle = 16\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.
    \kern-\nulldelimiterspace} 4}} {\left[ { - \frac{1}{{2{{\cos }^2}\phi }}\int\limits_0^1 {\sqrt {1 - {\rho ^2}{{\cos }^2}\phi } d\left( {1 - {\rho ^2}{{\cos }^2}\phi } \right)} } \right]d\phi } = $

    $\displaystyle = 16\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.
    \kern-\nulldelimiterspace} 4}} {\left[ {\left. {\left( { - \frac{1}
    {{3{{\cos }^2}\phi }}} \right)\sqrt {{{\left( {1 - {\rho ^2}{{\cos }^2}\phi } \right)}^3}} } \right|_{\rho = 0}^{\rho = 1}} \right]d\phi } = $

    $\displaystyle = \frac{{16}}{3}\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.\kern-\nulldelimiterspace} 4}} {\frac{{1 - {{\sin }^3}\phi }}
    {{{{\cos }^2}\phi }}} d\phi = \frac{{16}}{3}$$\displaystyle \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.\kern-\nulldelimiterspace} 4}} {\left[ {\frac{1}{{{{\cos }^2}\phi }} - \frac{{\left( {1 - {{\cos }^2}\phi } \right)\sin \phi }}{{{{\cos }^2}\phi }}} \right]} d\phi =
    $

    $\displaystyle = \frac{{16}}{3}\left[ {\left. {\tan \phi } \right|_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.\kern-\nulldelimiterspace} 4}} + \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 4}} \right.
    \kern-\nulldelimiterspace} 4}} {\frac{{1 - {{\cos }^2}\phi }}
    {{{{\cos }^2}\phi }}} d\left( {\cos \phi } \right)} \right] =$

    $\displaystyle = \frac{{16}}{3}\left[ {1 + \int\limits_0^{{\pi \mathord{\left/
    {\vphantom {\pi 4}} \right.\kern-\nulldelimiterspace} 4}} {\left( {\frac{1}
    {{{{\cos }^2}\phi }} - 1} \right)d\left( {\cos \phi } \right)} } \right] =$

    $\displaystyle = \frac{{16}}{3}\left[ {\left. {1 - \left( {\frac{1}
    {{\cos \phi }} + \cos \phi } \right)} \right|_0^{{\pi \mathord{\left/
    {\vphantom {\pi 4}} \right.\kern-\nulldelimiterspace} 4}}} \right] = \frac{{16}}{3}\left[ {1 - \left( {\frac{{3\sqrt 2 }}{2} - 2} \right)} \right] =$

    $\displaystyle = \frac{{16}}{3}\left[ {3 - \frac{{3\sqrt 2 }}{2}} \right] = 16 - 8\sqrt 2 = 8\left( {2 - \sqrt 2 } \right) \approx 4.686$
    Last edited by DeMath; Feb 16th 2009 at 09:00 AM.
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  4. #4
    Senior Member DeMath's Avatar
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    Look this graphic interpretation

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