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Math Help - Area problem

  1. #1
    Senior Member
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    Area problem

    Determine the area in the 1st quadrant bounded by x=acos^3t, y=asin^3t, the x-axis and the y-axis (a>0)

    In my solution book it says:

    As x varies from 0 to a, t varies from \frac {\pi} {2} to 0

    How do you know that?

    Skipping all the way down to this step

    3a^2\int_\frac{\pi} {2} ^2 sin^4tdt - 3a^2\int_\frac{\pi} {2} ^2 sin^6tdt

    = [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{\pi} {2}] - [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{5} {6}*\frac{\pi} {2}]

    How do you get to that step?

    Thanks a lot.
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  2. #2
    Member Mentia's Avatar
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    Bellingham, WA
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    So what they did was transform to a coordinate system that is easier to integrate.

    You can just plug in values to see what happens to x as t varies, it shows that as x varies 0->a, t varies 0->Pi/2.

    We know x=aCos(t)^{3}, so

    \left | \frac  {dx} {dt} \right | =3aCos(t)^{2}Sin(t), this is our Jacobian.

    Then we have:

    Area =  \int_{ 0}^{ \frac{  \pi }{2 }  }  \int_{0 }^{aSin(t)^{3} } 3aCos(t)^{2}Sin(t)dydt = \int_{0}^{ \frac{\pi}{2}} 3a^{2} \cos ^2(t) \sin ^4(t) dt

    But we have cos(t)^2 = 1 - sin(t)^2 which gives:

    \int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^4(t)dt - \int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^6(t)dt

    I'm not 100% sure why their limits are Pi/2 to 2 instead of my 0 to Pi/2 but it doesn't seem to change the answer.
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  3. #3
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    I've never learned double integrals before.

    I know I'm supposed to plug in values from 0 to a, but what values? What is the limit of a? (a<1?, a<2? a<3?)

    I get this step

    <br />
3a^2\int_\frac{\pi} {2} ^2 sin^4tdt - 3a^2\int_\frac{\pi} {2} ^2 sin^6tdt <br />

    But I don't get the step after that.
    Last edited by chengbin; March 9th 2009 at 04:01 AM.
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