# Where is the error?

1. Mar 26, 2015

### Schwarzschild90

1. The problem statement, all variables and given/known data

2. Relevant equations

And the volume element dV expressed in spherical coordinates.

3. The attempt at a solution

Can you spot where I've made an error?

Last edited: Mar 26, 2015
2. Mar 26, 2015

### SammyS

Staff Emeritus
What's you question?

Oh! I see.

Usually the question is posted in the text of the thread.

3. Mar 26, 2015

### Schwarzschild90

Where is the error? * (Where have I made an error)

-Ok! Edited*

4. Mar 26, 2015

### SammyS

Staff Emeritus
I don't see the error after going through what you did twice.

5. Mar 26, 2015

### Ray Vickson

Why do you think you have made an error?

6. Mar 26, 2015

### haruspex

There is a slightly easier way. By symmetry you can change the x to z in the integrand.

7. Mar 27, 2015

### Schwarzschild90

My study friends have gotten a different result than I. (Pi/8). And my intuition tells me that by integrating over radius=1, I should get something that looks like 1/8 the volume of a sphere of radius 1 (Pi/6). Would someone like to clear this up? I'd like to know if I'm doing something wrong.

Last edited: Mar 27, 2015
8. Mar 27, 2015

### haruspex

Sure but the integrand, x, is on average less than 1/2.
You can check your answer by looking up the mass centre of a hemisphere.

9. Mar 27, 2015

### Schwarzschild90

I turned in the assigment. Giving it another shot tonight.

10. Mar 27, 2015

### Ray Vickson

Do it the easy way: $V = \int_0^1 x A(x) \, dx$, where $A(x)$ is the area of the slice at $x$ parallel to the $yz$ plane. The slice is a quarter circle of radius $\sqrt{1-x^2}$, so $A(x) = \frac{1}{4} \pi (1-x^2)$. This gives $V = \pi/16$, as you obtained.