Question

I just need to know how to solve for the area of the drawing at the bottom comment

Answer 1

It's significantly easier that way.

Answer 2

so i find for the intersection right?

\[2y - y^2 = y^2 - 4y\]
\[2y^2 - 6y = 0\]
\[2y(y - 3) = 0\]
y = 0
y = 3

so these are my upper and lower limits

\[\int_{0}^{3} (2y - y^2 -[y^2-4y])dy\]

did i do it right?

Answer 3

Yes, perfect.

Answer 4

yay thanks :DD i think i got the curves down \m/ btw...how do you do those circles?? they're weird

Answer 5

Hmm? Circles?

Answer 6

i dont know my teacher showed examples having those buggers

Answer 7

Is that what you're askin' for?

Answer 8

example