# area integral problem

• September 17th 2007, 06:53 PM
chocolatelover
area integral problem
Hi everyone,

Can somone tell me if this is correct, please?

I'm trying to find the area under the curves x=ysquared-4y and x=2y-ysquared

I took 2y-y^2 -y^2-4ydy from -3 to 0.

I got 62/2

Thank you very much
• September 17th 2007, 08:29 PM
CaptainBlack
Quote:

Originally Posted by chocolatelover
Hi everyone,

Can somone tell me if this is correct, please?

I'm trying to find the area under the curves x=ysquared-4y and x=2y-ysquared

I took 2y-y^2 -y^2-4ydy from -3 to 0.

I got 62/2

Thank you very much

Area under or between?

The two curves intersect where $y^2-4y=2y-y^2$, to me the
points of intersection appear to be $y=0,\ y=3$, so you want:

$
\int_0^3 (2y-y^2)-(y^2-4y)\ dy = \int_0^3 6y-2y^2\ dy
$

RonL
• September 17th 2007, 08:40 PM
chocolatelover
Thank you very much

I got 9 as my final answer. Does that look right?

Thank you
• September 17th 2007, 09:52 PM
AfterShock
Quote:

Originally Posted by chocolatelover
Thank you very much

I got 9 as my final answer. Does that look right?

Thank you

Yes, this is correct.