MHB Spherical co-ordinates conversion

[Fermat1]

Consider the region above the $z=-\sqrt{2-x^2-y^2}$ and below $z=-\sqrt{x^2+y^2}$.

Let $x=r\sin\phi\cos\theta$, $y=r\sin\phi\sin\theta$, $z=r\cos\phi$

I want the range of the variables. I get $0\leq r\leq\sqrt{2}$.

How do I work out the range of $\phi$ and $\theta$ ?

 

[I like Serena]

Consider the region above the $z=-\sqrt{2-x^2-y^2}$ and below $z=-\sqrt{x^2+y^2}$.

Let $x=r\sin\phi\cos\theta$, $y=r\sin\phi\sin\theta$, $z=r\cos\phi$

I want the range of the variables. I get $0\leq r\leq\sqrt{2}$.

How do I work out the range of $\phi$ and $\theta$ ?

Simplest is by drawing those surfaces.
The first is a demi-sphere below the $z=0$ plane.
The second is a cone, also below the $z=0$ plane.

In polar coordinates you will have the full range for the longutide $0 \le \theta < 2\pi$.

Edited:
And the colatitude begins at the cone.
That is, $\frac{3\pi}{4} \le \phi \le \pi$.

 

[Fermat1]

Simplest is by drawing those surfaces.
The first is a demi-sphere below the $z=0$ plane.
The second is a cone, also below the $z=0$ plane.

In polar coordinates you will have the full range for the longutide $0 \le \theta < 2\pi$.
And the colatitude begins at the $z=0$ plane and ends at the cone.
That is, $\frac \pi 2 \le \phi \le \frac{3\pi}{4}$.

Hi can you explain how you found those angles-how do they relate to the equations for z?Thanks

 

[I like Serena]

Hi can you explain how you found those angles-how do they relate to the equations for z?Thanks

Let's take a look at the cone $z=-\sqrt{x^2+y^2}$.

Substituting the definitions of the polar coordinates, we get:
$$r\cos\phi = -\sqrt{(r\sin\phi\cos\theta)^2 + (r\sin\phi\sin\theta)^2} = -r\sin\phi$$
Since $\theta$ is not present in this equation, we need its full range.
Solving it for $\phi$ in the domain $[0,\pi]$ gives $\phi=\frac{3\pi}{4}$.

So to integrate above the cone, we need $0 \le \theta < 2\pi$ and $\phi \le \frac{3\pi}{4}$.

 

[Fermat1]

Let's take a look at the cone $z=-\sqrt{x^2+y^2}$.

Substituting the definitions of the polar coordinates, we get:
$$r\cos\phi = -\sqrt{(r\sin\phi\cos\theta)^2 + (r\sin\phi\sin\theta)^2} = -r\sin\phi$$
Since $\theta$ is not present in this equation, we need its full range.
Solving it for $\phi$ in the domain $[0,\pi]$ gives $\phi=\frac{3\pi}{4}$.

So to integrate above the cone, we need $0 \le \theta < 2\pi$ and $\phi \le \frac{3\pi}{4}$.

thanks, do you mean integrate below the cone?

 

[I like Serena]

thanks, do you mean integrate below the cone?

Ah. To integrate below the cone it should be $\phi \ge \frac{3\pi}{4}$.

 

[Fermat1]

Ah. To integrate below the cone it should be $\phi \ge \frac{3\pi}{4}$.

How do you get the upper limit? When I use the other equation I 'lose' ${\phi}$. Does that signify there is no upper restriction (apart from ${\pi}$)?

 

[I like Serena]

How do you get the upper limit? When I use the other equation I 'lose' ${\phi}$. Does that signify there is no upper restriction (apart from ${\pi}$)?

Yes.

 

[Fermat1]

Yes.

How about this. Calculate the outward flux through the surface $x^2+y^2+z^2=1$ of a fluid with velocity field $F$.

When I convert to spherical, the range of both anges should be [0,${\pi}$/2] Why is there this restriction?

Thanks

 

[I like Serena]

How about this. Calculate the outward flux through the surface $x^2+y^2+z^2=1$ of a fluid with velocity field $F$.

When I convert to spherical, the range of both anges should be [0,${\pi}$/2] Why is there this restriction?

Thanks

There is no implicit reason for such a restriction.
The text must have explicitly stated any restrictions.

 

[Fermat1]

There is no implicit reason for such a restriction.
The text must have explicitly stated any restrictions.

see here homework - Finding the outward flux through a sphere - Mathematics Stack Exchange

Cheers

 

[I like Serena]

see here homework - Finding the outward flux through a sphere - Mathematics Stack Exchange

Cheers

That problem statement states explicitly: "in the first octant".
It restricts both angles to $[0,\frac \pi 2]$.

Cheers