Find volume of a solid figure that lies between: $x^2+y^2+z^2\leq 4$, $2x^2+2y^2-2z^2\geq 1$, $2z^2\geq x^2+y^2$ Find volume of a solid figure that lies between:
$x^2+y^2+z^2\leq 4$,
$2x^2+2y^2-2z^2\geq 1$,
$2z^2\geq x^2+y^2$
I just really can't figure out the limits of integration... any hint would be great
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Spherical coordinates:
$x=\rho \sin\varphi \cos\theta$
$y=\rho \sin\varphi \sin\theta$
$z = \rho \cos\varphi$
From sphere: $\rho ^2 \leq 4$ or $\rho \leq 2$.
From hyperboloid: $ 2\rho ^2( \sin^2\varphi - \cos^2 \varphi)\geq 1$
From cone: $\rho ^2(2 \cos^2\varphi - \sin^2 \varphi)\geq 0$
Intersection of sphere and hyperboloid, I got the ellipse:
$\frac{x^2}{\frac{9}{4}}+\frac{y^2}{\frac{9}{4}}=1$ or in spherical coordinates: $\rho^2 \sin^2\varphi=\frac{9}{4}$
 A: Let's consider the region above $z = 0$. Due to symmetry the volume bound is same above and below $z = 0$.
The region is defined by,
a) $x^2 + y^2 + z^2 \leq 4$
In spherical coordinates, $\rho \leq 2$ (the sphere)
b) $2x^2+2y^2-2z^2\geq 1$
In spherical coordinates, $ - 2 \rho^2 \cos 2\phi\geq 1$
c) $2 z^2 \geq x^2 + y^2$
In spherical coordinates, $ \tan \phi \leq \sqrt2$
At intersection of hyperboloid and sphere,
$-8 \cos2\phi = 1$ (plugging in $\rho = 2$ in $(b)$)
So, $\cos 2\phi = - \frac{1}{8}$
Notice that this is the region which is outside the hyperboloid and inside the cone and the sphere.
So, $ \sqrt{- (\sec 2\phi) / 2} \leq \rho \leq 2$ (please note for the given limits of $\phi$, $\sec 2\phi$ is negative and so the value inside the square root is positive).
$\arccos \left(\frac{\sqrt7}{4}\right) \leq \phi \leq \arccos \left(\frac{1}{\sqrt3}\right)$ (the lower bound is same as $\cos 2 \phi \leq - 1/8$ and the upper bound is same as $\tan \phi \leq \sqrt2$. I have just rewritten them differently).
$0 \leq \theta \leq 2\pi$
Can you take it from here?
