The discussion centers on calculating the area of a shaded region defined by the polar equation r = sqrt(θ). The correct integral for the area is A = 1/2 ∫ from 3π/2 to 2π of r² dθ, which simplifies to A = 1/2[(ln(2π) - ln(3π/2)]. The confusion arose from misinterpretation of the integrand, with some participants mistakenly suggesting the integral should be 1/(2r²). Ultimately, the correct limits of integration were confirmed as 3π/2 and 2π.
PREREQUISITESStudents studying calculus, particularly those focusing on polar coordinates, as well as educators looking for examples of common misunderstandings in integral calculus.
Hi BWJ,Bigworldjust said:Lol, honestly I am new to this forum, so I wasn't even familiar with LaTeX :P. Next time, I'll be sure to use it tho.
Off my back? Have I offended you?sharks said:Heh. At least something positive came out of it, and it'll get oay off my back.
oay said:Hi BWJ,
Even though LaTeX is great for writing out maths, it certainly wasn't necessary for your question.
Did you get your answer to be...\frac{7}{16}\pi^2
No probs. Mathematicians always argue. But it's always me who is right.Bigworldjust said:Yeah that's the answer I got :D. Thanks for the help everyone, didn't mean to spark such a debate, hah.