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Thread: area of disc

  1. #16
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    Re: area of disc

    ok thanks. how did you know it was this?
    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
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  2. #17
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    Re: area of disc

    Quote Originally Posted by markosheehan View Post
    ok thanks. how did you know it was this?
    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:

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