Find the volume of the solid which is restricted to the sphere $\rho=1$ and cones $\varphi=\pi/6$ and $\varphi=\pi/3$.
I always use a grapher to see how exactly the solid looks like,but in this case the surfaces are described in spherical coordinates and I don't know any program which plots them, so I need to convert them in Cartesian coordinates system,can someone help?
I know that $\rho^2=1=x^2+y^2+z^2$ and so we have a sphere of radius $1$, but what about the cones? How to convert them?
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Let's say we have a cone in spherical coordinates with the equation $\phi = \phi_0$.
From the definitions of our spherical coordinates we have $z = \rho \cos\phi$, and $0 \leq \phi \leq \pi$, so $\phi = \arccos\left(\dfrac{z}{\rho}\right).$
Now we have $\arccos\left(\dfrac{z}{\rho}\right) = \phi_0,$ so converting everything to Cartesian we have $\dfrac{z}{\sqrt{x^2 + y^2 + z^2}}=\cos(\phi_0).$ Now square both sides: $$\dfrac{z^2}{x^2 + y^2 + z^2} = \cos^2 (\phi_0) \to z^2 = \cos^2(\phi_0)(x^2 + y^2 + z^2) \to \sin^2(\phi_0) z^2 = \cos^2(\phi_0)(x^2 + y^2)$$.
Dividing both sides by $\sin^2(\phi_0)$ gives us $z^2 = \cot^2(\phi_0)(x^2 + y^2)$.
The volume bound between cones and sphere is straightforward in spherical coordinates. In this case for example, bounds of $\rho$ and $\theta$ are defined by the sphere. The lower and upper bounds of $\phi$ is defined by the cone. But if you had to convert this into cartesian coordinates, which is your question,
In spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin \phi, z = \rho \cos\phi$.
$\sqrt {x^2+y^2} = \rho \sin\phi, z = \rho \cos\phi, \tan \phi = \displaystyle \frac {\sqrt{x^2+y^2}}{z}$
For cone $ \ \phi = \frac{\pi}{6}$, $\displaystyle \tan \frac{\pi}{6} = \frac {\sqrt{x^2+y^2}}{z}$
So the cone in cartesian coordinates is $z = \sqrt {3 \ (x^2+y^2)}$
Find similarly for cone $\phi = \frac{\pi}{3}$