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Thread: Surface area of parametric curve

  1. #1
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    Surface area of parametric curve

    The curve defined by x = a cos3 t, y = a sin3 t for 0 ≤ t ≤ 2π, a ≠ 0, is known as an astroid.

    The curve could be approximated by the arcs of four quadrants each of radius a.
    By considering the coordinates of particular points on the curve and on the approximating arcs of the quadrants for corresponding values of t, determine whether the total length of the astroid is greater or less than the sum of the arcs of the quadrants.

    Find the total (exact) length of the asteroid by integration.
    Hence confirm the approximation described above.
    Establish the integral required to find the surface area formed when a parametric curve
    x = f(t), y = g(t) is rotated about the x-axis.

    Find the surface area formed when the curve x = a cos3 t, y = a sin3 t (0 ≤ t ≤ π/2) is rotated about the x-axis.

    Compare the surface area obtained with the surface area of a hemisphere of radius a.
    From your work in part a), why might some people expect the surface area found above to be similar to the surface area of a hemisphere, and why are they actually rather different?
    Last edited by user654321; Apr 27th 2012 at 12:47 PM.
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  2. #2
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    Re: Surface area of parametric curve

    Length of the approximation is the perimeter of the circle with radius a, hence the perimeter is calculated as $\displaystyle l_{approx}=2a\pi\approx 6.28 a$.

    The exact perimeter via integration is obtained by calculating the length of part of the astroid in the first quadrant multiplied by 4 (symmetry). Astroid in the first quadrant is obtained for parameter $\displaystyle t\in [0,\frac{\pi}{2}]$. So here goes:
    $\displaystyle l_{exact}=4\cdot \int\limits_{0}^{\frac{\pi}{2}}\sqrt{\left( x'(t) \right) ^2+\left(y'(t)\right)^2}dt=4\cdot \int\limits_{0}^{\frac{\pi}{2}}\sqrt{\left(3a \cos^2 t \sin t \right)^2+\left(3a \sin ^2 t \cos t\right)^2}dt=$
    $\displaystyle =4\cdot \int\limits_{0}^{\frac{\pi}{2}}\sqrt{9a^2 \cos^4 t \sin^2 t +9 a^2 \sin ^4 t \cos^2 t}dt=4\cdot \int\limits_{0}^{\frac{\pi}{2}}\sqrt{9a^2 \cos^2 t \sin^2 t \left( \cos^2 t +\sin^2 t \right)}dt=$
    $\displaystyle =4\cdot \int\limits_{0}^{\frac{\pi}{2}}3a\cos t \sin t dt=12a\int\limits_{0}^{\frac{\pi}{2}} \sin t \, d\left(\sin t\right)=12a \left(\frac{\sin ^2 t}{2}\right)\bigg\vert_{0}^{\frac{\pi}{2}}=6a.$

    So you have the difference between the approximation and astroid $\displaystyle \vert 6.28a-6a \vert=0.28a$ - for any given $\displaystyle a$ the error in perimeter of the approximation is always roughly $\displaystyle 0.28a$.
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  3. #3
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    Re: Surface area of parametric curve

    As for the surface one can easily find the formula: $\displaystyle S_x=2\pi\int\limits_{a}^{b} y(t)\cdot \sqrt{\left( x'(t)\right) ^2 + \left( y'(t)\right) ^2}dt$.

    Similar to above, calculate only one half of the surface and multiply it by two. If you focus on the half on the right of the y-axis then $\displaystyle S_x=2\cdot 2\pi\int\limits_{0}^{\frac{\pi}{2}} y(t)\cdot \sqrt{\left( x'(t)\right) ^2 + \left( y'(t)\right) ^2}dt$.
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