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.

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- Aug 2nd 2006, 10:06 AMbret80Area
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. - Aug 2nd 2006, 12:00 PMQuickQuote:

Originally Posted by**bret80**

but if you want to know how to solve this problem, you'll have to wait for someone else. - Aug 2nd 2006, 01:19 PMThePerfectHackerQuote:

Originally Posted by**bret80**

You need to isolate the point where they intersect.

$\displaystyle \frac{2x^2}{3}=-x^2+15$

Thus,

$\displaystyle 2x^2=-3x^2+45$

Thus,

$\displaystyle 5x^2=45$

Thus,

$\displaystyle x^2=9$

Thus,

$\displaystyle x=\pm 3$

Thus, the integral is from -3 to 3.

Now you need to subtract the lower curve from the top curve.

Thus,

$\displaystyle \int_{-3}^3 -x^2+15-\frac{2}{3}x^2 dx$

Thus,

$\displaystyle \int_{-3}^3 -\frac{5}{3}x^2+15dx$

Integrate,

$\displaystyle -\frac{5}{9}x^3+15x\big|^3_{-3}$

Is,

$\displaystyle -\frac{5}{9}3^3+15(3)+\frac{5}{9}(-3)^3-15(-3)$=$\displaystyle -15+45-15+45=60$