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Thread: Volume of a Spatial Figure

  1. #1
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    Volume of a Spatial Figure

    Please, help me to decide this task

    Find the volume of a spatial figure which is limited the implicate function:

    $\displaystyle \left( {\frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} + \frac{{z^2 }}{{c^2 }}} \right)^2 = $$\displaystyle \frac{z}{h}\exp \left( {\frac{{{{x^2 } \mathord{\left/{\vphantom {{x^2 } {c^2 }}} \right.\kern-\nulldelimiterspace} {c^2 }}}}{{{{x^2 } \mathord{\left/{\vphantom {{x^2 } {a^2 }}} \right.\kern-\nulldelimiterspace} {a^2 }} + {{y^2 } \mathord{\left/{\vphantom {{y^2 } {b^2 }}} \right.\kern-\nulldelimiterspace} {b^2 }} + {{z^2 } \mathord{\left/{\vphantom {{z^2 } {c^2 }}} \right.\kern-\nulldelimiterspace} {c^2 }}}}} \right),$

    $\displaystyle a,b,c,h{\text{ }} - {\text{ positive parameters}}{\text{.}}$

    Please, say: which a system of coordinates I must use?
    I tried to use the cylindrical system of coordinates, but a very difficult triple integral.
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  2. #2
    Senior Member vincisonfire's Avatar
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    My guess would be spherical coordinates (slightly modified).
    Try
    $\displaystyle x = a\cdot r\cdot sin(\phi)cos(\theta)$
    $\displaystyle y = b\cdot r\cdot sin(\phi)sin(\theta)$
    $\displaystyle z = c\cdot r\cdot cos(\phi) $
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  3. #3
    Senior Member DeMath's Avatar
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    I used also the generalized spherical coordinates.
    There's my solution here.

    So, we need find a volume $\displaystyle V$ of area $\displaystyle G \subset \mathbb{R}^3$, which is limited a surface $\displaystyle S$:

    $\displaystyle S = \left\{ {\left. {x,y,z} \right|\left( {\frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} + \frac{{z^2 }}{{c^2 }}} \right)^2 } \right.$$\displaystyle \left. { = \frac{z}{h}\exp \left( {\frac{{{{x^2 } \mathord{\left/{\vphantom {{x^2 } {c^2 }}} \right. \kern-\nulldelimiterspace} {c^2 }}}}{{{{x^2 } \mathord{\left/{\vphantom {{x^2 } {a^2 }}} \right.\kern-\nulldelimiterspace} {a^2 }} + {{y^2 } \mathord{\left/{\vphantom {{y^2 } {b^2 }}} \right.\kern-\nulldelimiterspace} {b^2 }} + {{z^2 } \mathord{\left/{\vphantom {{z^2 } {c^2 }}} \right.\kern-\nulldelimiterspace} {c^2 }}}}} \right)} \right\}.$

    Obviously, $\displaystyle V = \iiint\limits_G {dV}.$

    There's a using of the generalized spherical coordinates:

    $\displaystyle \left\{ \begin{gathered}x = ar\cos \phi \sin \theta , \hfill \\y = br\sin \phi \sin \theta , \hfill \\z = cr\cos \theta . \hfill \\ \end{gathered} \right.$

    Where $\displaystyle r > 0$, $\displaystyle 0 < \theta < \pi$, $\displaystyle 0 < \phi < 2\pi$.

    It's jacobian of the transformation:

    $\displaystyle J = \left| {\begin{array}{*{20}c}{\frac{{\partial x}}{{\partial r}}} & {\frac{{\partial x}}{{\partial \phi }}} & {\frac{{\partial x}}{{\partial \theta }}} \\{\frac{{\partial y}}{{\partial r}}} & {\frac{{\partial y}}{{\partial \phi }}} & {\frac{{\partial y}}{{\partial \theta }}} \\{\frac{{\partial z}}{{\partial r}}} & {\frac{{\partial z}}{{\partial \phi }}} & {\frac{{\partial z}}{{\partial \theta }}} \\\end{array} } \right| = $$\displaystyle ^{} \left| {\begin{array}{*{20}c}{a\cos \phi \sin \theta } & { - ar\sin \phi \sin \theta } & {ar\cos \phi \cos \theta } \\{b\sin \phi \sin \theta } & {br\cos \phi \sin \theta } & {br\sin \phi \cos \theta } \\{c\cos \theta } & 0 & { - cr\sin \theta } \\\end{array} } \right| = $$\displaystyle ^{} abcr^2 \sin \theta {\text{ }}\left( {{\text{absolute value}}} \right).$

    $\displaystyle \left( {\frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} + \frac{{z^2 }}{{c^2 }}} \right)^2 = \left( {\frac{{\left( {ar\cos \phi \sin \theta } \right)^2 }}{{a^2 }} + \frac{{\left( {br\sin \phi \sin \theta } \right)^2 }}{{b^2 }} + \frac{{\left( {cr\cos \theta } \right)^2 }}{{c^2 }}} \right)^2 = $

    $\displaystyle = \left( {r^2 \cos ^2 \phi \sin ^2 \theta + r^2 \sin ^2 \phi \sin ^2 \theta + r^2 \cos ^2 \theta } \right)^2 $$\displaystyle = r^4 \left( {\sin ^2 \theta \left( {\cos ^2 \phi + \sin ^2 \phi } \right) + \cos ^2 \theta } \right)^2 = r^4 .$

    $\displaystyle r^4 = \frac{{cr\cos \theta }}{h}\exp \left( {\frac{{a^2 r^2 \cos ^2 \phi \sin ^2 \theta }}{{c^2 r^2 }}} \right) \Leftrightarrow r^3 = \frac{c}{h}\cos \left( \theta \right)e^{\left( {\frac{a}{c}} \right)^2 \sin ^2 \left( \theta \right)\cos ^2 \left( \phi \right)} .$

    So, we have [$\displaystyle 0 < \theta < \frac{\pi }{2}$ because $\displaystyle r>0$]

    $\displaystyle V = abc\int\limits_0^{2\pi } {d\phi \int\limits_0^{{\pi \mathord{\left/
    {\vphantom {\pi 2}} \right.\kern-\nulldelimiterspace} 2}} {\sin \left( \theta \right)d\theta \int\limits_0^{R\left( {\phi ,\theta } \right)} {r^2 } dr{\text{ }}} } $ where $\displaystyle R\left( {\phi ,\theta } \right) = \sqrt[3]{{\frac{c}{h}\cos \left( \theta \right)e^{\left( {\frac{a}{c}} \right)^2 \sin ^2 \left( \theta \right)\cos ^2 \left( \phi \right)} }}.$

    The calculation of the internal integrals.

    $\displaystyle \int\limits_0^{R\left( {\phi ,\theta } \right)} {r^2 } dr = \left. {\frac{{r^3 }}{3}} \right|_0^{R\left( {\phi ,\theta } \right)} = \frac{c}{{3h}}\cos \left( \theta \right)e^{\left( {\frac{a}{c}} \right)^2 \sin ^2 \left( \theta \right)\cos ^2 \left( \phi \right)} .$

    $\displaystyle \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 2}} \right.\kern-\nulldelimiterspace} 2}} {\sin \theta d\theta } \int\limits_0^{R\left( {\phi ,\theta } \right)} {r^2 } dr = \frac{c}{{3h}}\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 2}} \right.\kern-\nulldelimiterspace} 2}} {\sin \theta \cos \left( \theta \right)e^{\left( {\frac{a}{c}} \right)^2 \sin ^2 \left( \theta \right)\cos ^2 \left( \phi \right)}d\theta } = $$\displaystyle \left\{ \begin{gathered}\sin ^2 \left( \theta \right) = t, \hfill \\2\sin \left( \theta \right)\cos \left( \theta \right)d\theta = dt \hfill \\ \end{gathered} \right\} = $

    $\displaystyle = \frac{c}{{6h}}\int\limits_0^1 {e^{t\left( {\frac{a}{c}} \right)^2 \cos ^2 \left( \phi \right)} dt} = \left. {\frac{c}{{6h}}\left( {\frac{a}{c}\cos \phi } \right)^{ - 2} e^{t\left( {\frac{a}{c}\cos \phi } \right)^2 } } \right|_0^1 = \frac{c}{{6h}}\left[ {\frac{{e^{\left( {\frac{a}{c}\cos \phi } \right)^2 } - 1}}{{\left( {\frac{a}{c}\cos \phi } \right)^2 }}} \right] = $

    $\displaystyle = \frac{c}{{6h}}\left( {\frac{a}{c}\cos \phi } \right)^{ - 2} \left[ {\left( {\frac{a}{c}\cos \phi } \right)^2 + \frac{1}{{2!}}\left( {\frac{a}{c}\cos \phi } \right)^4 + \ldots + \frac{1}{{n!}}\left( {\frac{a}{c}\cos \phi } \right)^{2n} + \ldots} \right] = $

    $\displaystyle = \frac{c}{{6h}}\left[ {1 + \frac{1}{{2!}}\left( {\frac{a}{c}\cos \phi } \right)^2 + \ldots + \frac{1}{{n!}}\left( {\frac{a}{c}\cos \phi } \right)^{2n - 2} + \ldots } \right] = \frac{c}{{6h}}\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}\left( {\frac{a}{c}\cos \phi } \right)^{2n - 2} .} $

    So, we have

    $\displaystyle V = abc\int\limits_0^{2\pi } {d\phi \int\limits_0^{{\pi \mathord{\left/
    {\vphantom {\pi 2}} \right.\kern-\nulldelimiterspace} 2}} {\sin \left( \theta \right)d\theta \int\limits_0^{R\left( {\phi ,\theta } \right)} {r^2 } dr = } } \frac{{abc^2 }}{{6h}}\int\limits_0^{2\pi } {\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}\left( {\frac{a}{c}\cos \phi } \right)^{2n - 2} } d\phi } = $$\displaystyle \frac{{abc^2 }}{{6h}}\sum\limits_{n = 1}^\infty {\left[ {\frac{1}{{n!}}\left( {\frac{a}{c}} \right)^{2n - 2} \int\limits_0^{2\pi } {\left( {\cos \phi } \right)^{2n - 2} d\phi } } \right].} $

    The integral $\displaystyle \int\limits_0^{2\pi } {\left( {\cos \phi } \right)^{2n - 2} d\phi }$, calculated by the methods of complex analysis, is equal $\displaystyle \frac{{2\pi \left( {2n - 2} \right)!}}{{2^{2n - 2} \left( {\left( {n - 1} \right)!} \right)^2 }}$.

    Finally, we have

    $\displaystyle V = \frac{{\pi abc^2 }}{{3h}}\sum\limits_{n = 1}^\infty {\frac{{\left( {2n - 2} \right)!}}{{2^{2n - 2} \left( {\left( {n - 1} \right)!} \right)^2 n!}}\left( {\frac{a}{c}} \right)^{2n - 2} } .$
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