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Find the area of region enclosed by astroid

  1. Apr 15, 2011 #1
    1. The problem statement, all variables and given/known data

    find the region enclosed by astroid

    x= acosϑ 3
    y=asinϑ 3

    the astroid looks like this picture in my book, with the shaded region stretching for x (-a,a)

    and for y(-a,a)


    2. Relevant equations

    i think the formula is ..... from (-a to a) ∫ y * x' dϑ

    3. The attempt at a solution

    a = ∫ acosϑ 3 * 3asinϑ 2 cosϑ

    a = 3a2∫ cosϑ 4 * sinϑ 2

    a = 3a2∫ cosϑ 4 * (1-cosϑ 2 )

    a = 3a2∫ cosϑ 4 - cosϑ6

    is this one the right track, if so I can evaluate the rest of that integral myself
  2. jcsd
  3. Apr 16, 2011 #2

    I like Serena

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    Your formula would be:

    [tex]A=\int_{-a}^a y dx = \int_{\pi}^0 y \frac {dx} {d\theta} d\theta[/tex]

    Please note that this will be only the area above the x-axis.
    To get the total area, you will need to double that.

    Your derivative of x should be:

    [tex]\frac {dx} {d\theta}=\frac {d} {d\theta} (a\cos^3\theta)=3a\cos^2\theta\cdot -\sin\theta[/tex]

    Substituting gives:

    [tex]A=\int_{\pi}^0 (a\sin^3\theta) \cdot (3a\cos^2\theta\cdot -\sin\theta) d\theta
    =3a^2\int_0^{\pi} \sin^4\theta \cos^2\theta d\theta

    I think you can take it from there.
    So you were on the right track, but made a mistake in substitution.

    [edit]Also, you should take note of the integral boundaries. They can not just be left out.[/edit]
  4. Apr 18, 2011 #3
    quick question, how did you decide the bound from pi to 0 and then you flipped them, it has something to do with the negative sign right
  5. Apr 18, 2011 #4

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    In the first equation we shift from x to theta.
    A value of x=-a corresponds to a value theta=pi, and x=+a corresponds to theta=0.
    So the integral is from pi to 0.

    In the third equation, there is a minus sign in the equation.
    Flipping the boundaries means flipping the sign, or in this case removing the minus sign.
  6. Apr 18, 2011 #5
    okay i was scratching my head about this one but i think you get it by just substituting x for a and solving for theta right?

    -a = acos3ϑ

    -1 = cos3ϑ

    ϑ = pi

    a = acos3ϑ

    1 = cos3ϑ

    ϑ = 0

    did i arrive at that correctly?
  7. Apr 18, 2011 #6

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    Yes. That is entirely correct! :smile:

    Actually, what I did is look at the graph you posted.
    With theta being the angle with the positive x-axis, a value of pi corresponds to x=-a.
    Substituting pi in the expression for x shows that this is correct.
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