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The image below shows a circle of radius r with the bottom right quadrant arc in red. Is there an equation for this and only this arc? The reason for the question: what is the area under that arc, highlighted in green? The easy answer is:

$$\frac{4r^2-\pi r^2}4$$

I'm looking for an answer using integral calculus.

enter image description here

José Carlos Santos
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charlie_sar
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  • When you studied calculus you might have seen the integral motivated by a picture showing the area under a curve. What equation describes the arc of this circle? – Ethan Bolker Aug 28 '22 at 18:10
  • that's the question I'm asking. I know the equation for the circle. – charlie_sar Aug 28 '22 at 18:13
  • @JoséCarlosSantos - I was trying mathjax, unknown user error, terribly sorry. Do you not understand the question? – charlie_sar Aug 28 '22 at 18:19
  • I have edited the formula. Did I get it right? – José Carlos Santos Aug 28 '22 at 18:33
  • You say “I know the equation for the circle.” How about providing it and showing how you think it might be used in an integral to express the area? In other words, show what you’ve tried, as is the norm for asking questions here. – Steve Kass Aug 28 '22 at 18:36
  • the equation for this unit circle would be x^2 + y^2 -2y = 0. I've found several sources like https://www.analyzemath.com/calculus/Integrals/area_circle.html to use integration to find the area of a circle, but not to answer the question posed. – charlie_sar Aug 28 '22 at 18:53

1 Answers1

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The answer is standard. The circle has equation

$(y-r)^{2}+x^{2}=r^{2}$ which gives:

$y^{2}-2ry+x^{2}=0$. Solving for $y$ we get $y=r\pm\sqrt{r^{2}-x^{2}}$.

Since $y(0)=0$ we have $y=r-\sqrt{r^{2}-x^{2}}$ and the area is

$\int_{0}^{r}(r-\sqrt{r^{2}-x^{2}})dx$. Using standard integration methods

(setting $x=rsin\theta$) we obtain that:

$A=r^{2}-\dfrac{r^{2}\pi}{4}$