# Volume integral in cylindrical coordinates

1. Oct 10, 2015

### whatisreality

1. The problem statement, all variables and given/known data
OK, I thought once I knew what the question was asking I'd be able to do it. I was wrong!

Consider the volume V inside the cylinder x2 +y2 = 4R2 and between z = (x2 + 3y2)/R and the (x,y) plane, where x, y, z are Cartesian coordinates and R is a constant. Write down a triple integral for the volume V using cylindrical coordinates. Include the limits of integration (three upper and three lower). Evaluate the integral to determine the volume V in terms of R.

2. Relevant equations

3. The attempt at a solution
I've sketched both equations, so I have a vague idea of what's going on. The z=... equation is an 'elliptic paraboloid', the other eqn a cylinder. I could find the volume by finding the volume of the cylinder and subtracting the volume above the paraboloid, although the limits for that might be complicated. That's what I'd do in 2D, never done a 3D question.

I think I have to find the z coordinate where they intersect, but I don't know how. And I'm supposed to convert to cylindrical, which is probably the only bit I can at least try, using x=rcos(θ), y=rsin(θ) and z=z:

r2cos2(θ) + r2sin2(θ) = 4R2
And
(r2cos2(θ) + 3r2sin2(θ) )/R = z

I think I could replace r with 2R? Then the equations become
cos2(θ) + sin2(θ) = 1
And
4Rcos2(θ) + 12Rsin2(θ)=z
Not sure where to go now!

Last edited: Oct 10, 2015
2. Oct 10, 2015

### Ray Vickson

Your integration region $R$ is 3-dimentisonal:
$$R = \{ (x,y,z): x^2 + y^2 \leq 4 R^2, 0 \leq z \leq (x^2 + 3 y^2)/R$$
In cylindrical coordinates $x = r \cos \theta, y = r \sin \theta$ these become
$$\begin{array}{l} r^2 \leq 4 R^2 \Longrightarrow 0 \leq r \leq 2 R \\ 0 \leq z \leq r^2 (\cos^2 \theta + \sin^2 \theta + 2 \sin^2 \theta)/R = r^2 (1 + 2 \sin^2 \theta)/R \\ 0 \leq \theta \leq 2 \pi \end{array}$$

Last edited: Oct 10, 2015
3. Oct 10, 2015

### whatisreality

OK, I can follow that. And then the function I actually perform the volume integral on would be the z=... function, with those limits?

4. Oct 10, 2015

### Ray Vickson

You tell me.

5. Oct 10, 2015

### whatisreality

Those are definitely the limits of the integration. And the function to be integrated obviously has to be in cylindrical coordinates too. So I think the function should be the z= function in cylindrical coordinates.

6. Oct 10, 2015

### whatisreality

Because if I integrated the other function, that would just give me the volume of the cylinder. But I'm still not really confident with my answer!

Last edited: Oct 10, 2015
7. Oct 10, 2015

### Ray Vickson

I am not sure what your issue is here. What was the answer you got? What detailed formulas did you use to get it? Did you do it two different ways and get two different answer? What, exactly, is the problem?

8. Oct 10, 2015

### whatisreality

Oh, no, haven't computed the integral yet. My issue was, initially, I didn't know what to do, which I now do. The answer I'm not confident with is the answer that I should be integrating the z= function, even though I'm nearly 100% sure it's right. But I'm nearly never confident in my answers, so that's fine! :)

I do know how to actually do the integral now that I have the limits though. Triple integral with respect to r dr dθ dz. I really appreciate you taking the time to reply by the way, thank you for your help!

9. Oct 11, 2015

### whatisreality

Well, now I have done the integral, and I do actually still have a problem...

So since V=∫∫∫ r dr dθ dz, you integrate r dr dθ dz with the limits found?
Although actually, you have to rearrange the z= function to get an expression for r. Which would be

r = $\sqrt{ \frac{zR}{1+2 \sin^2(\theta)}}$

Which I would not enjoy integrating. Or do you just make a function f(r, θ, z) by turning z = $\frac{r^2(1+2\sin^2(\theta))}{R}$ into

f(r, θ, z) =$\frac{r^2(1+2\sin^2(\theta))}{R} - z$? I don't actually know what to integrate.

10. Oct 11, 2015

### whatisreality

Er, not just ∫∫∫ r dr dθ dz, I think that might be just for cylinders actually.