# multiple integrals

• Feb 4th 2009, 07:18 PM
Andreamet
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:
• Feb 5th 2009, 10:18 AM
Opalg
Google "Steinmetz solid" for information and proofs.
• Feb 5th 2009, 08:50 PM
DeMath
Quote:

Originally Posted by Andreamet
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

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

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

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

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

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

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

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

$= \frac{{16}}{3}\left[ {3 - \frac{{3\sqrt 2 }}{2}} \right] = 16 - 8\sqrt 2 = 8\left( {2 - \sqrt 2 } \right) \approx 4.686$
• Feb 7th 2009, 01:45 PM
DeMath
Look this graphic interpretation