Area Inside Circle x^2+y^2=a^2 Above b=7

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Discussion Overview

The discussion revolves around finding the area inside the circle defined by the equation x² + y² = a², specifically the area that lies above the line y = b, where b is constrained by -a ≤ b ≤ a. Participants explore different interpretations of the problem and methods for calculating the area.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant suggests calculating the area by integrating from -a to a and subtracting the area from -a to b.
  • Another participant questions whether the original problem statement contains a typo, proposing that it should refer to "above y = b" instead of "above 7 = b".
  • A later reply indicates a misunderstanding of the initial remark and seeks clarification on the area calculation.
  • Another participant proposes that a geometric approach might be simpler than using integrals, referencing a method involving the area of a triangle.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the problem statement and the method of solution. There is no consensus on the interpretation of the question or the best approach to find the area.

Contextual Notes

Participants note potential ambiguities in the problem statement, particularly regarding the interpretation of "above 7 = b" and whether it should be "above y = b".

wonguyen1995
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Find area inside circle x^2+y^2=a^2, above 7=b, -a \le b \le a ?? i think f of y right?
(-a to a) minus (-a to b)?
 
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wonguyen1995 said:
Find area inside circle x^2+y^2=a^2, above 7=b, -a \le b \le a
Maybe it's a typo and should read "above $y=b$". That is, find the area inside the circle $x^2+y^2=a^2$ that is located above the line $y=b$, where $-a\le b\le a$.

wonguyen1995 said:
?? i think f of y right?
(-a to a) minus (-a to b)?
Sorry, I don't understand your remark.
 
Evgeny.Makarov said:
Maybe it's a typo and should read "above $y=b$". That is, find the area inside the circle $x^2+y^2=a^2$ that is located above the line $y=b$, where $-a\le b\le a$.

Sorry, I don't understand your remark.

That is it, i mistake.
so can you help me??
 
Do you need a solution that uses integral? It's easier to find the area geometrically as described in Wikipedia using the fact that the area of a triangle with sides $u$ and $v$ and angle $\varphi$ between them is $\frac{1}{2}uv\sin\varphi$.
 

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