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.
My computer program says that the shaded area in the graph below is 60 (although that number is probably rounded)...Originally Posted by bret80
but if you want to know how to solve this problem, you'll have to wait for someone else.
First look at Quick's beautiful graph.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,
$\displaystyle -\frac{5}{9}3^3+15(3)+\frac{5}{9}(-3)^3-15(-3)$=$\displaystyle -15+45-15+45=60$