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Math Help - 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:
    {x^2} + {y^2} = 1;
    {y^2} + {z^2} = 1;
    {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 OZ; second cylinder axis coincides with the axis OX; third cylinder axis coincides with the axis OY. We have a symmetry, so we should calculate the amount of body parts in the first oktant, and the answer multiplied by 8. Next, we consider the case x \geqslant y. Then the upper limit of integration for z we can determine from the equation {z^2} + {x^2} = 1 which gives us a lower value for z, other than equation {z^2} + {y^2} = 1. The field of integration D in the plane XOY, we can determine from the equation {x^2} + {y^2} = 1, i.e. this area is 1/8 of unit circle.

    So we have V = 16\iint\limits_D {dxdy}\int\limits_0^{\sqrt {1 - {x^2}} } {dz}  = 16\iint\limits_D {\sqrt {1 - {x^2}} dxdy}, 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

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

    = 16\int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  4}} \right.<br />
 \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 }  =

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

    = \frac{{16}}{3}\int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  4}} \right.\kern-\nulldelimiterspace} 4}} {\frac{{1 - {{\sin }^3}\phi }}<br />
{{{{\cos }^2}\phi }}} d\phi  = \frac{{16}}{3} \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  =<br />

    = \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.<br />
 \kern-\nulldelimiterspace} 4}} {\frac{{1 - {{\cos }^2}\phi }}<br />
{{{{\cos }^2}\phi }}} d\left( {\cos \phi } \right)} \right] =

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

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

    = \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; February 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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