The discussion revolves around a double integral involving symmetry in the context of evaluating integrals over semi-circular regions in the xy-plane. Participants are examining the implications of symmetry on the limits of integration and the evaluation of the integral itself.
The discussion is ongoing, with participants exploring the reasoning behind the use of symmetry and the evaluation of the integral. Some guidance has been offered regarding the careful evaluation of square roots, but there is no explicit consensus on the correctness of the integral in question.
Participants are navigating potential misunderstandings about the limits of integration and the properties of square roots, indicating a need for clarity on these assumptions in the context of the problem.
ainster31 said: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 said: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.
You have to be careful whenever you 'execute' a square root. √(x2) is not the same as x.ainster31 said:Alright, but why is the integral shown in the remark incorrect?