# Thread: Area enclosed by a graph and a chord

1. ## 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?

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

Area bounded by chord - Area bounded by the curve

3. 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.

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

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

5. 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!