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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find the area between the graphs you'll have to first rewrite 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't have to rewrite it in terms of y. Keep them as they are.. find the intersection points, that's: \[2y^2=12y^2 \implies y=2, y=2\] If you draw them, you will see that 12y^2 is "on the right" of 2y^2 in the given interval, so your area will be: \[A=\int\limits\limits_{2}^{2}(12  y^2 2y^)dy =\int\limits\limits_{2}^{2}(123y^2) dy=12yy^3=32\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense to you, Ebrake?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea when i did it out I got 32 but i just wanted to make sure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey sstarica, Where have you been? long time no see.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, here and there, everywhere? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're the one whose been out of sight lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL.. I guess you're right :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm good, how abt you ^_^?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am fine.. Thanks for asking ^^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
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