# Triple Integral in Cylindrical Coordinates

1. Nov 10, 2008

### daveyman

1. The problem statement, all variables and given/known data
Find the mass and center of mass of the solid S bounded by the paraboloid $$z=4x^2+4y^2$$ and the plane $$z=a\;\;(a>0)$$ if S has constant density K.

2. Relevant equations
In cylindrical coordinates, $$x^2+y^2=r^2$$.

3. The attempt at a solution
In order to find the mass, I tried
$$\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K\;drdzd\theta$$
This didn't seem to work, though.

2. Nov 10, 2008

### tiny-tim

Hi daveyman!

Hint: the volume element isn't drdzdθ.

3. Nov 10, 2008

### daveyman

Okay, so this is the mass:

$$\int _0^{2\pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}\text{Kr}\;\;drdzd\theta = \frac{1}{8} a^2 K \pi$$

Last edited: Nov 10, 2008
4. Nov 10, 2008

### daveyman

Now that I have the mass (m), I want to find the r component of the center of mass. I tried

$$\frac{1}{m}\int _0^{2 \pi }\int _0^a\int _0^{\frac{\sqrt{z}}{2}}K r^2drdzd\theta$$

This is incorrect. What am I doing wrong?

Last edited: Nov 10, 2008
5. Nov 10, 2008

### tiny-tim

That should come out as 0, which is the r component of the centre of mass.

6. Nov 10, 2008

### daveyman

Mathematica is telling me that this integral is not equal to zero. Have I made a mistake?
I've included the output as an attachment.

#### Attached Files:

• ###### Output.pdf
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48
7. Nov 10, 2008

### tiny-tim

ah … just realised … you can't do centre of mass that way in cylindrical coordinates … you'd have to convert to x and y coordinates first.

(but the z coordinate will be ok )

8. Nov 10, 2008

### daveyman

Right, that makes sense. So how can I solve for the r and theta components in cylindrical coordinates? I know qualitatively that they should be zero, but how do I show this mathematically?

9. Nov 10, 2008

### tiny-tim

I think you're perfectly entitled to say that it's obvious.

(btw, if r is zero, then θ is meaningless )

10. Nov 10, 2008

### daveyman

Very true. I think that's it. Thank you!

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