1. ## Re: area of disc

so from here x=asin(t) ....
then dx=a*-cos(t) dt still not exactly sure why this is.
now i am trying to find the limits so when x=a ... t=90 so

∫ 90,-90 2* √[b^2 - ((b^2 *a^2sin(t)^2)/a^2) a*-cos(t) dt ]

i am not rally sure how to integrate this

2. ## Re: area of disc

Originally Posted by markosheehan
so from here x=asin(t) ....
then dx=a*-cos(t) dt still not exactly sure why this is.
now i am trying to find the limits so when x=a ... t=90 so

∫ 90,-90 2* √[b^2 - ((b^2 *a^2sin(t)^2)/a^2) a*-cos(t) dt ]

i am not rally sure how to integrate this
The derivative of $a\sin t$ is $a\cos t dt$, not $a(-\cos t) dt$.

If you factor out the $b^2$ inside the square root, you get: $b^2\left(1-\left(\dfrac{x}{a}\right)^2\right)$. My goal is to get rid of the square root. So, I am looking for a substitution that will yield a perfect square. The factor of $b^2$ is already a perfect square, and I know that $1-\sin^2 t = \cos^2 t$. So, I want $\dfrac{x}{a} = \sin t$. I could have just as easily used $\dfrac{x}{a} = \cos t$ since $1-\cos^2 t = \sin^2 t$ is also a perfect square.

The integral becomes:

$\displaystyle 2b\int_{-\tfrac{\pi}{2}}^\tfrac{\pi}{2}\sqrt{1-\sin^2 t}a\cos t dt = 2ab\int_{-\tfrac{\pi}{2}}^\tfrac{\pi}{2} \sqrt{\cos^2 t}\cos t dt = 4ab\int_0^\tfrac{\pi}{2} \cos^2 t dt$

We use the double angle formula for cosine to simplify:
$\cos(2t) = 2\cos^2 t -1$
So, $\cos^2 t = \dfrac{\cos (2t) + 1}{2}$

Plugging in, we get:

$\displaystyle 2ab\int_0^\tfrac{\pi}{2}(\cos(2t)+1)dt = ab\left[\sin(2t) + 2t\right]_0^\tfrac{\pi}{2} = ab\left[\sin \pi + \pi - \sin 0 - 0\right] = ab\pi$

Page 2 of 2 First 12