Volume integration where the radius limit depends on the polar angle

[M_1]

Hi,
I want to calculate a volume integral of a function f(r,theta). The limits for azimuthal and polar angle is of course 0-2*pi and 0-pi, respectively. But the limits for the radius is 0 to an expression depending on the polar angle. Can I simply first integrate the r-part, say from r^4 to 1/5*r^5, and then put in the expression with the polar angle and then integrate?

See attached Word-file for explanations.

Many thanks!

 

[Astrum]

What do you mean by "outer radius"?

That's one ugly looking integral. What is this for?

 

[M_1]

Hi,

Thanks for your answer.

By outer radius I simply mean the radius of the droplet.

All this is for calculating the vibrational energy of a vibrating droplet. I you want I can send you a more complete document.

 

[Astrum]

I'm not really sure if what you're doing makes sense.

If you wanted to evaluate that integral, you need to sub in the expression for $r$. The limits of the $dr$ integral are just $[r,0]$.

It would look something like this:

$$\int ^{2\pi} _0 \int ^\pi _0 \int ^r _0 r^4 cos^2 (\theta)sin^3 (\theta) dr d\theta d\phi$$

And you need to plug in that hideous expression for $r$.

I'll be honest with you, the reason you haven't gotten many responses is because this looks like total nonsense.

A tip: Use the built in math script, nobody wants to download a file from a random user on an internet forum. It has a bit of a learning curve, but you'll get used to it.

 

[M_1]

Hi again,

Thanks!

The question is if I can integrate r⁴ to 1/5r⁵ and then put in the expression for r (Eq. 2).

Best Regards

 

[Astrum]

Certainly not, you need to sub the expression in before you can integrate anything.

 

[M_1]

But if I plug in the expression for r, how should I proceed with the r-integration. Should I see it as the function 1 and integrate to the primitive function r? This will differ by a factor of 1/5 as compared to to first integrate r⁴ to 1/5r⁵ and then put in the expression for r (Eq. 2).

 

[Astrum]

Plug the expression for $r$ in, and integrate as you normally would. For example

$$\int ^{2 \pi}_0 \int ^r _0 r dr d \theta$$

Where $r = krcos(\theta)$ would be rewritten as:

$$\int ^{2 \pi}_0 \int ^r _0 (kr)cos(\theta)dr d\theta$$

You just sub it in, and THEN do the integration.