Area enclosed by a graph and a chord

**ment2byours** · May 1st 2009, 07:13 PM

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?

**adhyeta** · May 1st 2009, 07:37 PM

since the chord is actually above the curve, you'll need to find out:

Area bounded by chord - Area bounded by the curve

**ment2byours** · May 1st 2009, 07:40 PM

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.

**redsoxfan325** · May 1st 2009, 07:44 PM

Originally Posted by ment2byours

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 $\int_{-1}^8 \frac{1}{3}x+\frac{4}{3}-x^{2/3}\,dx$

$= \left[\frac{x^2}{6}+\frac{4x}{3}-\frac{3}{5}x^{5/3}\right]_{-1}^8$ $= \frac{64}{6}+\frac{32}{3}-\frac{3}{5}\cdot 32 - \left(\frac{1}{6}+\frac{4}{3}-\frac{3}{5}\cdot-1\right) = \boxed{\frac{1}{30}}$

**ment2byours** · May 1st 2009, 07:56 PM

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!

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