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Math Help - I have to solve this equation or term in order to calculate the area

  1. #1
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    I have to solve this equation or term in order to calculate the area

    Hi, my teacher did the following step, and I cant understand it really:

    We have A = b $(With limits from 0 to a) Sqroot [1-x^2 / a^2 ] dx

    he got the common denominator as following: a^2 -x^2 / a^2

    Now he suddenly wrote it like this ....


    b/a $ Sqroot [a^2 - x^2] dx My queston is how did he do this step??
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  2. #2
    MHF Contributor harish21's Avatar
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    Do you mean \displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;?

    For \sqrt{\dfrac{a^2-x^2}{a^2}},

    Note: for every a>0, \sqrt{a^2} = (a^2)^{\frac{1}{2}}= a

    therefore, \sqrt{\dfrac{a^2-x^2}{a^2}}\;=\;\dfrac{\sqrt{a^2-x^2}}{a}\;=\; \dfrac{1}{a} \times \sqrt{a^2-x^2}

    so, \displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;=\; \dfrac{b}{a} \int_0^a \sqrt{a^2-x^2}\;dx
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  3. #3
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    I would have done that slightly differently. Let x= sin(\theta) so that dx= cos(\theta)d\theta and \sqrt{1-\frac{x^2}{a^2}}= \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta). When x= 0, sin(\theta)= 0 so that \theta= 0 and when x= a, sin(\theta)= 1 so that \theta= \pi/2. The integral becomes
    b\int_0^{\pi/2}cos^2(\theta)d\theta.

    Or, since this was posted in "Pre-algebra and Algebra", I might recognize that y= b\sqrt{1- \frac{x^2}{a^2} leads to y^2= b^2\left(1- \frac{x^2}{a^2}\right) or \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1, an ellipse which has are \pi ab. Since a square root is always positive, this is the upper portion of that area and has area \frac{\pi ab}{2}.
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