Question

Find the area between the graphs of x=2y^2 and x=12-y^2

Answer 1

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

Answer 2

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:
\[A=\int\limits\limits_{-2}^{2}(12 - y^2 -2y^)dy =\int\limits\limits_{-2}^{2}(12-3y^2) dy=12y-y^3=32\]

Answer 3

Does that make sense to you, Ebrake?

Answer 4

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

Answer 5

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

Answer 6

lol, here and there, everywhere? :)

Answer 7

You're the one whose been out of sight lol

Answer 8

LOL.. I guess you're right :)

Answer 9

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

Answer 10

I am fine.. Thanks for asking ^^

Answer 11

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