Area

**bret80** · August 2nd 2006, 10:06 AM

Find the area of the region limited by the graphs of

y= 2x^2/3 and y= -x^2+15

Nice Graph by Quick, but I would like to know how to calculate the result, thanks.

**Quick** · August 2nd 2006, 12:00 PM

Originally Posted by bret80

Find the area of the region limited by the graphs of

y= 2x^2/3 and y= -x^2+15

My computer program says that the shaded area in the graph below is 60 (although that number is probably rounded)...
but if you want to know how to solve this problem, you'll have to wait for someone else.

**ThePerfectHacker** · August 2nd 2006, 01:19 PM

Originally Posted by bret80

Find the area of the region limited by the graphs of

y= 2x^2/3 and y= -x^2+15

First look at Quick's beautiful graph.

You need to isolate the point where they intersect.
$\frac{2x^2}{3}=-x^2+15$
Thus,
$2x^2=-3x^2+45$
Thus,
$5x^2=45$
Thus,
$x^2=9$
Thus,
$x=\pm 3$
Thus, the integral is from -3 to 3.
Now you need to subtract the lower curve from the top curve.
Thus,
$\int_{-3}^3 -x^2+15-\frac{2}{3}x^2 dx$
Thus,
$\int_{-3}^3 -\frac{5}{3}x^2+15dx$
Integrate,
$-\frac{5}{9}x^3+15x\big|^3_{-3}$
Is,
$-\frac{5}{9}3^3+15(3)+\frac{5}{9}(-3)^3-15(-3)$ = $-15+45-15+45=60$

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