Area of a Circle: Solving the Equation

[Vector1962]

TL;DR

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.

 

[PeroK]

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 $x$.

 

[Mark44]

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.

 

[malawi_glenn]

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 $t = \pi / 2$

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

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