Question

Integrating a circle...

Answer 1

For fun, I'm trying to integrate the area of a circle in hopes of getting A = pi r^2.
I started with the function of a circle:\[(x-h)^{2}+(y-k)^{2} = r^{2}\] I can assume h and k are 0, then I solved for y:\[y = \sqrt{r^{2} - x^{2}}\] Really this is the function of the top half/semicircle of a circle, so to find the area of the circle, I need to double its integration:\[A = 2\int\limits \sqrt{r^{2} - x^{2}}dx\] Using trig substitution, I get\[A = 2r\int\limits \cos \theta \sqrt{r^{2} - r^{2}\sin^{2} \theta}d \theta\]\[A = 2r^{2}\int\limits \cos^{2} \theta d \theta \]\[A = r^{2}\int\limits (1 + \cos2 \theta) d \theta \]\[A = r^{2}\theta + \frac{1}{2}r^{2}\sin2 \theta\] \[A = r^{2}\sin^{-1}\frac{x}{r} + \frac{1}{2}r^{2}\sin(2\sin^{-1}\frac{x}{r}) \]\[A = r^{2}\sin^{-1}\frac{x}{r} + rx \cos(\sin^{-1}\frac{x}{r}) \]\[A = r^{2}\sin^{-1}\frac{x}{r} + x \sqrt{r^{2} - x^{2}}\]
And now I'm stuck. Any suggestions?

Answer 2

Oh wait, do I need to integrate it from -r to r?

Answer 3

using polar coordinates would be a lot easier, but I suppose you are avoiding that on purpose

Answer 4

^^ to answer your question, yes.

Answer 5

if you change the bounds using your trig sub to -pi/2 to pi/2 this is really not so bad

Answer 6

you don't have to undo you whole sub that way

Answer 7

the second term cancels out in both cases. arcsin(1) = pi/2, and arcsin(-1) = -pi / 2.

Answer 8

\[\theta \in \left[ 0, \pi \right]\] in our case

Answer 9

\[-r=r\sin\theta\to\theta=-\frac{\pi}2\]\[r=r\sin\theta\to\theta=\frac{\pi}2\]\[A = r^{2}\theta + \frac{1}{2}r^{2}\sin2 \theta|_{-\pi/2}^{\pi/2}=\pi r^2\]

Answer 10

No, if I change my bounds, it's from -pi/2 to pi/2, so \[[r^{2}(\pi/2) + 2r^{2}\sin(\pi/2)\cos(\pi/2)] - [r^{2}(-\pi/2) + 2r^{2}\sin(-\pi/2)\cos(-\pi/2)] \]\[r^{2}(\pi/2) + 2r^{2}(1)(0) + r^{2}(\pi/2) - 2r^{2}(-1)(0) = \pi r^{2}\] YAY

Answer 11

lol yay!