Question

Integral de volumes = elipsóide
[tex] 2x^{2} + y^{2} \leq 1      

Answer 1

Sea $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ un elipsoide, hallaremos su volumen.

1) usaremos el siguiente cambio de variable
$x=ar\cos \theta\sin \phi\;,\; y = br\sin \theta\sin \phi\;,\; z=cr\cos \phi$

donde: $\theta \in (0,2\pi]\;,\; \phi\in (0,\pi]\;,\;r\in(0,1]$

2) Cálculo del jacobiano

      $\left|\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi )}\right|=\left| \det\left[\begin{matrix} a\cos\theta\sin \phi&-ar\sin\theta\sin \phi&ar\cos\theta\cos \phi\\ b\sin\theta\sin\phi&br\cos\theta\sin\phi&br\sin\theta\cos\phi\\ c\cos\phi&0&-cr\sin\phi \end{matrix}\right]\right|\\ \\ \\ \left|\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi )}\right|=abcr^2|\sin \phi|$

3) Calculo del volumen

              $\displaystyle V=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}abcr^2|\sin\phi|\,dr\,d\phi\,d\theta\\ \\ V=abc\int_{0}^{2\pi}d\theta\cdot \int_{0}^{\pi}|\sin \phi|d\phi\cdot \int_{0}^{1}r^2dr\\ \\ V=2\pi \,abc\cdot \int_{0}^{\pi}\sin \phi \;d\phi\cdot \dfrac{1}{3}\\ \\ V=\dfrac{2}{3}\pi \,abc \cdot 2\\ \\ \\ \boxed{V=\dfrac{4}{3}\pi \,abc} \\ \\.$