Quick question about finding area for polar coordinates

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Homework Help Overview

The discussion revolves around finding the area of a shaded region defined by the polar equation r = sqrt(θ). Participants are exploring the appropriate limits of integration for the area calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants question the correct values for the limits of integration, with suggestions to consider the quadrant in which the shaded area lies. Others discuss the relationship between r and θ and the implications for setting up the integral.

Discussion Status

There is an ongoing exploration of the correct parameters for the integral, with some participants providing insights into the relationship between r and θ. Misinterpretations of the integrand have been noted, but clarity is sought regarding the correct setup for the area calculation.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of direct solutions provided. There is a noted confusion regarding the integrand, with some participants clarifying the correct form of the area integral.

  • #31
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.
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
 
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  • #32
sharks said:
Heh. At least something positive came out of it, and it'll get oay off my back.
Off my back? Have I offended you?
 
  • #33
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

Yeah that's the answer I got :D. Thanks for the help everyone, didn't mean to spark such a debate, hah.
 
  • #34
Bigworldjust said:
Yeah that's the answer I got :D. Thanks for the help everyone, didn't mean to spark such a debate, hah.
No probs. Mathematicians always argue. But it's always me who is right.
 
Last edited:

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