The discussion focuses on evaluating the double integral of the function f(x,y) = xy over the region defined by x ≥ 0, y ≥ 0, and x² + y² ≤ 4 using polar coordinates. Participants clarify the correct limits for the integral, emphasizing that the upper limit for r should be 2, not 2sec(θ), as the boundary is a circle of radius 2. The correct integral setup is confirmed as f^{π/2}_{0} f^{2}_{0} r² cos(θ) sin(θ) dr dθ, leading to the evaluation of the integral after applying u-substitution.
PREREQUISITESStudents and educators in calculus, particularly those focusing on integration techniques, as well as anyone looking to deepen their understanding of polar coordinates in multivariable calculus.
DrunkApple said:Homework Statement
f(x,y) = xy
x ≥ 0, y ≥ 0, x^{2} + y^{2} ≤ 4
Homework Equations
The Attempt at a Solution
f^{pi/2}_{0}f^{2secθ}_{0} 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:f^{pi/2}_{0}f^{2}_{0} rcosθ * rsinθ * r drdθ
So it's not this?