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Math Help - Area enclosed by a graph and a chord

  1. #1
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    Area enclosed by a graph and a chord

    Find the area of the region bounded by the graphs of y^3=x^2 and the chord joining the points (-1,1) and (8,4)

    so when i turned y=cubed root (x^2) i found out that the points of the chord are shared with the graph so my integral is from [-1,8]. Do I just find the integral of y=cubed root (x^2) from [-1,8] or is it a lil more complicated than that?
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  2. #2
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    since the chord is actually above the curve, you'll need to find out:

    Area bounded by chord - Area bounded by the curve
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  3. #3
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    How can I find the area of the chord from [-1,8]? I tried making up an equation for it, but it does not seem to work for me.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ment2byours View Post
    Find the area of the region bounded by the graphs of y^3=x^2 and the chord joining the points (-1,1) and (8,4)

    so when i turned y=cubed root (x^2) i found out that the points of the chord are shared with the graph so my integral is from [-1,8]. Do I just find the integral of y=cubed root (x^2) from [-1,8] or is it a lil more complicated than that?
    First, find the equation of the line: m = \frac{4-1}{8-(-1)} = \frac{1}{3}

    y-1=\frac{1}{3}(x-(-1)) \implies y= \frac{1}{3}x+\frac{4}{3}

    You want the area below y= \frac{1}{3}x+\frac{4}{3} and above y^3=x^2

    So your integral is \int_{-1}^8 \frac{1}{3}x+\frac{4}{3}-x^{2/3}\,dx

    = \left[\frac{x^2}{6}+\frac{4x}{3}-\frac{3}{5}x^{5/3}\right]_{-1}^8 = \frac{64}{6}+\frac{32}{3}-\frac{3}{5}\cdot 32 - \left(\frac{1}{6}+\frac{4}{3}-\frac{3}{5}\cdot-1\right) = \boxed{\frac{1}{30}}
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  5. #5
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    Thanks that really helps. Maybe I should not try to do Calculus at night when I'm not as focus as I can be. I can finish the rest. Thank you guys so much!
    Last edited by ment2byours; May 2nd 2009 at 07:22 AM. Reason: 1
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