## anonymous 5 years ago Find the area between the graphs of x=2y^2 and x=12-y^2

1. anonymous

To find the area between the graphs you'll have to first re-write the given equations in terms of y :) 1) Sketch both functions in a graph 2) Find the intersection points when y1 = y2 --> this will yeild you the intervals ^_^ 3) Then subtract the curve that's on the top from the curve that's on the bottom; you can find out which is on the top and which is on the bottom from the graph. General equationf of finding the area between the graphs is : $A = \int\limits_{a}^{b} g(x) - f(x) dx$ where g(x) is the curve that's in the top, and f(x) is the curve that's in the bottom ^_^ I hope this helps because it'll take me ages to solve the problem :) give it a one more try

2. anonymous

You don't have to re-write it in terms of y. Keep them as they are.. find the intersection points, that's: $2y^2=12-y^2 \implies y=2, y=-2$ If you draw them, you will see that 12-y^2 is "on the right" of 2y^2 in the given interval, so your area will be: $A=\int\limits\limits_{-2}^{2}(12 - y^2 -2y^)dy =\int\limits\limits_{-2}^{2}(12-3y^2) dy=12y-y^3=32$

3. anonymous

Does that make sense to you, Ebrake?

4. anonymous

yea when i did it out I got 32 but i just wanted to make sure.

5. anonymous

Hey sstarica, Where have you been? long time no see.

6. anonymous

lol, here and there, everywhere? :)

7. anonymous

You're the one whose been out of sight lol

8. anonymous

LOL.. I guess you're right :)

9. anonymous

I'm good, how abt you ^_^?

10. anonymous

I am fine.. Thanks for asking ^^

11. anonymous

I got around 42.6 unless I read the integral wrong: A=∫−22(12−y2−2y)dy = ∫−22(12y−(1/3y^3)−y^2)dy = [12y−(1/3y^3)−y^2] from -2 to 2 then you just subsitute in 2 for y and substitute in -2 for y and then subtract one from the other and you get around 42.6