Forced to use symmetry to solve this double integral?

[ainster31]

Homework Statement

http://i.imgur.com/d4ViHux.png

Homework Equations

The Attempt at a Solution

The author writes: "Now, using symmetry, we have..."

But what symmetry does the author use? Also, I got the integral as shown in the remark but why is it wrong?

 

[cepheid]

Homework Statement

http://i.imgur.com/d4ViHux.png

Homework Equations

The Attempt at a Solution

The author writes: "Now, using symmetry, we have..."

But what symmetry does the author use? Also, I got the integral as shown in the remark but why is it wrong?

The symmetry that's used is only to integrate over only one semi-circle in the xy-plane, and not the other. That's why θ only ranges from 0 to π/2, not all the way to π. That's also why there is a factor of 2 in front of the integral: because the integral over one semi-circle should be exactly equal to the integral over the other one. This is because the portion of the hemisphere is that is above one semi-circle is equal in volume to the portion that is above the other: one portion is just the reflection of the other one across the y-axis. That is the symmetry.

 

[ainster31]

The symmetry that's used is only to integrate over only one semi-circle in the xy-plane, and not the other. That's why θ only ranges from 0 to π/2, not all the way to π. That's also why there is a factor of 2 in front of the integral: because the integral over one semi-circle should be exactly equal to the integral over the other one. This is because the portion of the hemisphere is that is above one semi-circle is equal in volume to the portion that is above the other: one portion is just the reflection of the other one across the y-axis. That is the symmetry.

Alright, but why is the integral shown in the remark incorrect?

 

[vela]

What did you get when you evaluated it? The integral isn't incorrect, but you have to be careful when evaluating it to get the correct result.

 

[haruspex]

Alright, but why is the integral shown in the remark incorrect?

You have to be careful whenever you 'execute' a square root. √(x²) is not the same as x.