The discussion revolves around evaluating a double integral in polar coordinates for the function f(x,y) = xy, constrained by x ≥ 0, y ≥ 0, and the circular boundary x² + y² ≤ 4.
The conversation is ongoing, with participants exploring different interpretations of the integration limits and expressing uncertainty about their approaches. Some participants have offered insights regarding potential overcomplications in the setup.
There is confusion regarding the correct interpretation of the boundary conditions, particularly whether the limits for r should be from 0 to 2 or involve a more complex expression. The discussion also touches on the implications of u-substitution in the context of the integral.
DrunkApple said:Homework Statement
f(x,y) = xy
x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4
Homework Equations
The Attempt at a Solution
[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ
I just wanted to check... is this right?? because I really don't think it is
right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?Dick said:Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
DrunkApple said:right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?
OH OH WAIT I JUST HAD AN EPIC EPIPHANY! THIS IS U SUBSTITUTION!
DrunkApple said:[itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
So it's not this?