How do spherical coordinates work for finding volume in a given region?

[mrcleanhands]

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.

Homework Equations

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

 

[HallsofIvy]

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.

Homework Equations

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].<br /> z= 1 gives. as you say, tan(\phi)= 3.<br /> <br /> [/quote]Assuming I did everything right I&#039;ll then get:<br /> E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\} <br /> <br /> Which I then plug into an integral like this:<br /> <br /> \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]<br /> Why do you still have x^2+ y^2 after you have converted to spherical coordinates?

 

[mrcleanhands]

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

 

[mrcleanhands]

bump^

 

[LCKurtz]

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?

 

[mrcleanhands]

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}

In post #1 you have shown that $\tan^2\phi = 3$, so $\tan \phi = \pm\sqrt 3$. If you plug that into a calculator you get $\phi = \pm \frac \pi 3$. Those are the principle values of the arctangent. But in spherical coordiantes you always want solutions between $0$ and $\pi$. So if you want the bottom portion of the cone, you need the second quadrant value of $\phi$ whose tangent is $-\sqrt 3$. That is $\frac{2\pi} 3$. Of course, you don't need that for your problem because it is only concerned with the upper half of the cone.