I Area of a Circle: Solving the Equation

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The discussion focuses on calculating the area of a circle using the equation x^2 + (y - r)^2 = r^2, with the center at (0, r). The original attempt to derive the area through integration resulted in an incorrect formula. Participants suggest that calculating the area of only a portion of the circle, such as half or a quarter, is more appropriate due to the multi-valued nature of y in the full circle equation. They emphasize the importance of showing the integral used to identify errors in the calculation. The conversation highlights the need for clarity in the integration process to achieve the correct area formula.
Vector1962
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area of a circle in terms of Y if center of circle is at (0 , r) --> A=f(y)
i can write the equation of circle easy enough, x^2+(y-r)^2=r^2. i get A=r^2/2 * asin((y-r)/r) + (y-r)/2 * sqrt(r^2 - (y-r)^2) through integration (using change of variable). Letting u = (y-r) and u^2=(y-r)^2, du= dy. Here's the rub... it's not right... :-) Appreciate and thanks in advance for any pointers... it's been a long time since I've done anything like this.
 
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Vector1962 said:
Summary: area of a circle in terms of Y if center of circle is at (0 , r) --> A=f(y)

i can write the equation of circle easy enough, x^2+(y-r)^2=r^2. i get A=r^2/2 * asin((y-r)/r) + (y-r)/2 * sqrt(r^2 - (y-r)^2) through integration (using change of variable). Letting u = (y-r) and u^2=(y-r)^2, du= dy. Here's the rub... it's not right... :-) Appreciate and thanks in advance for any pointers... it's been a long time since I've done anything like this.
I'm not sure I know what you are doing. Normally, you would calculate the area of half or a quarter of the circle using ##x^2 + (y - r)^2 = r^2##. If you try to do the whole circle, then ##y## is not a single-valued function of ##x##.
 
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Vector1962 said:
i get A=r^2/2 * asin((y-r)/r) + (y-r)/2 * sqrt(r^2 - (y-r)^2) through integration (using change of variable).
I agree with @PeroK's comment. Seeing only your result, but not the integral you used, it's hard to say why your result is wrong.
 
Here is a start for you

## x^2 + (y-r)^2 = r^2 ##

##u = y-r##

## \displaystyle \dfrac{A}{4} = \int_0^r \sqrt{r^2 - x^2}\mathrm{d} x
= r \int_0^r \sqrt{1 - x^2/r^2}\mathrm{d} x##

##x= r \sin t##, ##x = 0 ##gives ##t = 0##, ##x = r## gives ##t = \pi / 2 ##

##\mathrm{d}x = r \cos t \mathrm{d}t ##

## \displaystyle \dfrac{A}{4} = r\int_0^{\pi/2} \sqrt{1 - \sin^2t} \cos t \mathrm{d} t
##
 
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