# Spherical coordinates

## Homework Statement

Find $VR_{z}^{2} = \int \!\!\! \int \!\!\! \int_{E} (x^{2} + y^{2})dV$ given a constant density lying above upper half of $x^{2}+y^{2} = 3z^{2}$ and below $x^{2}+y^{2}+z^{2} = 4z$.

## The Attempt at a Solution

Why does it say upper half of $x^{2}+y^{2} = 3z^{2}$? It's not like there are two possible halves.

Here's what I've done so far but I'm a little unsure about some theory.
For $x^{2}+y^{2} = 3z^{2}$
$\tan^{2}\phi = 3$ which means $\phi=\pm\frac{\pi}{3}$

I've seen most questions don't actually have a - value for $\phi$ because it only works in a range of $0$ and $\pi$ right? The movement itself is taken care of by $\theta$? So I can just get rid off that negative $\frac{\pi}{3}$?

For $x^{2}+y^{2}+z^{2} = 4z$:
$\rho=4cos\phi$ and then usually questions will say $0\leq\rho\leq4\cos\phi$ but I don't really know how we get it ranges from 0... ?

Assuming I did everything right I'll then get:
$E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\}$

Which I then plug into an integral like this:

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta$

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HallsofIvy
Homework Helper

## Homework Statement

Find $VR_{z}^{2} = \int \!\!\! \int \!\!\! \int_{E} (x^{2} + y^{2})dV$ given a constant density lying above upper half of $x^{2}+y^{2} = 3z^{2}$ and below $x^{2}+y^{2}+z^{2} = 4z$.

## The Attempt at a Solution

Why does it say upper half of $x^{2}+y^{2} = 3z^{2}$? It's not like there are two possible halves.
On the contrary, there are "two possible halves", $z= \sqrt{(x^2+ y^2)/3}$ and $z= -\sqrt{(x^2+ y^2)/3}$. This is a cone with two nappes.

Here's what I've done so far but I'm a little unsure about some theory.
For $x^{2}+y^{2} = 3z^{2}$
$\tan^{2}\phi = 3$ which means $\phi=\pm\frac{\pi}{3}$

I've seen most questions don't actually have a - value for $\phi$ because it only works in a range of $0$ and $\pi$ right? The movement itself is taken care of by $\theta$? So I can just get rid off that negative $\frac{\pi}{3}$?
What negative are you talking about? Yes, in spherical coordinates, $\phi$ is always gbetween 0 and $\pi$. I don't see a problem with that.

For $x^{2}+y^{2}+z^{2} = 4z$:
$\rho=4cos\phi$ and then usually questions will say $0\leq\rho\leq4\cos\phi$ but I don't really know how we get it ranges from 0... ?
Where do the two surfaces, $x^2+ y^2= 3z^2$ (a cone) and $x^2+ y^2+ z^2= 4z$ (a sphere with center at (0, 0, 2)) intersect? Subtracting the first from the second we have $z^2= 4z- 3z^2$ which reduces to $z^2- z= z(z- 1)= 0[/tex]. z= 1 gives. as you say, [itex]tan(\phi)= 3$.

[/quote]Assuming I did everything right I'll then get:
$E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\}$

Which I then plug into an integral like this:

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta$[/QUOTE]
Why do you still have $x^2+ y^2$ after you have converted to spherical coordinates?

On the contrary, there are "two possible halves", $z= \sqrt{(x^2+ y^2)/3}$ and $z= -\sqrt{(x^2+ y^2)/3}$. This is a cone with two nappes.
Cool, I see it now.

What negative are you talking about?
This.

$x^{2}+y^{2}=3z^{2}$
$\rho^{2}\sin^{2}(\phi)\cos^{2}(\theta)+\rho^{2}\sin^{2}(\phi)\sin^{2}(\theta)=3\rho\cos^{2}(\phi)$
$\sin^{2}(\phi)=3\cos^{2}(\phi)$
$\tan^{2}\phi=3$
$\tan\phi=\pm\sqrt{3}$
$\phi=\pm\frac{\pi}{3}$

Where do the two surfaces, $x^2+ y^2= 3z^2$ (a cone) and $x^2+ y^2+ z^2= 4z$ (a sphere with center at (0, 0, 2)) intersect? Subtracting the first from the second we have $z^2= 4z- 3z^2$ which reduces to $z^2- z= z(z- 1)= 0$.
z= 1 gives. as you say, $tan(\phi)= 3$.
Ok. I see you've found the intersection using the rectangular coordinates.
Can't I do so in spherical co-ordinates?
Why am I only getting one of the intersections when I do it this way (spherical coordinates)? i.e. $\phi=\pm\frac{\pi}{3}$ and not the 0.
Also why is z=1 the equivalent of $\tan\phi = \pm\sqrt{3}$ ?
If I convert z=1 to spherical co-ordinates I get $\rho\sin\phi=1$ ... not sure how to convert the 1 to the spherical equivalent.
How did you get the centre of the sphere as (0,0,2)? The z on the other side of the equality has thrown me off a little.

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta$
Why do you still have $x^2+ y^2$ after you have converted to spherical coordinates?[/QUOTE]

Should be...

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (\rho^{2}\sin^{2}(\phi)) \,\rho^{2}\sin\phi d\rho d\phi d\theta$

Last edited:
bump^

LCKurtz
Homework Helper
Gold Member
Should be...

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (\rho^{2}\sin^{2}(\phi)) \,\rho^{2}\sin\phi d\rho d\phi d\theta$
bump^
Why the bump? What's left to do?

Yeah, I don't really get why z=1 gives $\tan\phi=\pm\sqrt{3}$ I can get this if I convert to spherical co-ordinates straight away but not by solving the intersection of the rectangular coordinates.

Also while there are no negative in spherical co-ordinate radians I get $\phi=\pm\frac{pi}{3}$

LCKurtz
Yeah, I don't really get why z=1 gives $\tan\phi=\pm\sqrt{3}$ I can get this if I convert to spherical co-ordinates straight away but not by solving the intersection of the rectangular coordinates.
Also while there are no negative in spherical co-ordinate radians I get $\phi=\pm\frac{pi}{3}$