The discussion focuses on the application of symmetry in solving a double integral over a semi-circle in the xy-plane. The author clarifies that the integration limits for θ should range from 0 to π/2, not π, due to the symmetry of the problem, which allows for the integration over one semi-circle while doubling the result to account for the other. Additionally, it is emphasized that the integral presented in the remark is not incorrect, but caution is necessary when evaluating square roots, as √(x²) does not equal x.
PREREQUISITESStudents studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of symmetry in mathematical problems.
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?