Triple Integration: Solve Homework Equation

[Quatros]

Homework Statement

I'm trying to figure out the other parameters to solve the problem

Homework Equations

I know sqrt(x^2+y^2+z^2) = p

The Attempt at a Solution

I changed the integrand to p^3 sin(theta) since p * p^2 sin(theta)
Then for the first integration sign, I know how to get the rest of them.

 

[Charles Link]

The differential volume $dV=r^2 \, \sin{ \theta} \, dr \, d \theta \, d \phi$ in spherical coordinates. You need to integrate the function over this and pick the limits on $r$, $\theta$, and $\phi$ correctly.

 

[Quatros]

The differential volume $dV=r^2 \, \sin{ \theta} \, dr \, d \theta \, d \phi$ in spherical coordinates. You need to integrate the function over this and pick the limits on $r$, $\theta$, and $\phi$ correctly.

but isn't that cylinderical cordinates.

 

[Charles Link]

but isn't that cylinderical cordinates.

No. Cylindrical coordinates is like polar coordinates in two dimensions and also has a $dz$.

 

[Quatros]

The problem asks for spherical coordinates. Read it carefully.

It's p for sphere our in our book, unless you aren't talking about integrand.

 

[Charles Link]

It's p for sphere our in our book, unless you aren't talking about integrand.

You can call it $p$ or $\rho$ or $r$, just so that you are consistent.

 

[Quatros]

Alright, I changed

Sqrt[x^2+y^2+z^2] to p * p^2 * sin(theta)

Would the z= sqrt(4-x^2+y^2) turn into sqrt(4-p^2)?

 

[Charles Link]

Where is the $dp$, $d \theta$ , and $d \phi$? Also, what do the limits need to be on $p$, $\theta$, and $\phi$? If you know how spherical coordinates work, the answer is simple. Incidentally, I think the book is using the Greek letter "rho"=$\rho$, but it's easier to use $r$.

 

[Quatros]

P 0 to sqrt(4-p^2)
Theta from 0 to 2pi?
the other symbol from 2 to 2?

 

[Charles Link]

The boundary of the surface you should recognize as a hemisphere of radius $r=2$. It is very easy to write the limits for it in spherical coordinates=Suggestion is you google "spherical coordinates" and study the explanation.

 

[Quatros]

The boundary of the surface you should recognize as a hemisphere of radius $r=2$. It is very easy to write the limits for it in spherical coordinates=Suggestion is you google "spherical coordinates" and study the explanation.

yeah, I did that and still confused. Sorry about my bad english.

 

[Charles Link]

yeah, I did that and still confused. Sorry about my bad english.

$\phi$ is the azimuth angle. It is the same angle as the $\phi$ in cylindrical coordinates. The $\theta$ is the tricky one in polar coordinates. It is measured from the z-axis. In the entire sphere, $\theta$ goes from 0 to $\pi$. For a hemisphere, it goes from $0$ to $\pi/2$. Polar coordinates is much easier explained holding a pencil and showing where it points. It's much easier to describe in person than in a write-up on the internet. But anyway, if you start with the pencil pointing upward, along the z-axis, that is $\theta =0$. Tilt the pencil down slightly, and you tilt it down by angle $\theta$. Depending on the direction you tilt it, that defines the $\phi$ angle...Rotate the tilted pencil (and yourself) around the z-axis (i.e. turn to the left or right), and $\theta$ stays constant, but you change $\phi$.

 

[Quatros]

Is this graph an Ice cream com

$\phi$ is the azimuth angle. It is the same angle as the $\phi$ in cylindrical coordinates. The $\theta$ is the tricky one in polar coordinates. It is measured from the z-axis. In the entire sphere, $\theta$ goes from 0 to $\pi$. For a hemisphere, it goes from $0$ to $\pi/2$. Polar coordinates is much easier explained holding a pencil and showing where it points. It's much easier to describe in person than in a write-up on the internet. But anyway, if you start with the pencil pointing upward, along the z-axis, that is $\theta =0$. Tilt the pencil down slightly, and you tilt it down by angle $\theta$. Depending on the direction you tilt it, that defines the $\phi$ angle...Rotate the tilted pencil (and yourself) around the z-axis (i.e. turn to the left or right), and $\theta$ stays constant, but you change $\phi$.

This is an ice cream cone, so the ball part (theta) is going from 0 to 2pi , P/R: 0 to 2,
The third parameter is giving me trouble.

 

[Charles Link]

Is this graph an Ice cream comThis is an ice cream cone, so the ball part (theta) is going from 0 to 2pi , P/R: 0 to 2,
The third parameter is giving me trouble.

(Referring to the hemisphere that is the boundary), $\phi$ goes from $0$ to $2 \pi$ ($360^o$). The polar angle $\theta$, goes from $0$ to $\pi/2$ ($90^o$). It's probably not obvious if this is the first time you used spherical coordinates, but with a little practice, it gets easier.

 

[Quatros]

(Referring to the hemisphere that is the boundary), $\phi$ goes from $0$ to $2 \pi$ ($360^o$). The polar angle $\theta$, goes from $0$ to $\pi/2$ ($90^o$). It's probably not obvious if this is the first time you used spherical coordinates, but with a little practice, it gets easier.

Is 8 pi correct?

 

[Charles Link]

Is 8 pi correct?

Yes, that's what I get also.