Thread:

1. Finding the area of this region

I've attached the problem I'm trying to solve. I know I'd want to set the two equal to find the points of intersection, but would I leave them in terms of X or Y? And then I would just take the integral of the top minus the bottom curve but I can't figure out WHAT is the top and bottom curve in this problem since they both kind of cross at the same point.
Attached Thumbnails

2. w/r to y ... (right curve) - (left curve)

$\int_0^5 (3y-y^2) - (y^2-7y) \, dy$

3. Ok my final answer was 112/3, and this may be a stupid follow up question, but would my unit just be units squared? It just seems weird to use units squared when dealing with odd shapes like this, but it's still correct right?

4. Originally Posted by fattydq
Ok my final answer was 112/3, and this may be a stupid follow up question, but would my unit just be units squared? It just seems weird to use units squared when dealing with odd shapes like this, but it's still correct right?
Yes, area is always in (linear units) squared.
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