Integrating to find area: x − 3y2 ≥ 0, 2 − x − 5|y| ≥ 0

[AmySaunders]

The problem is copied and pasted below.

I think the area I am trying to find is bound by x=3y^2 on the left, x=2-5y on the right, and the x axis. I then integrate from 0 to 1/3 (2-5y-3y^2)dy, and I get the answer 19/54. It seems so simple to me, but WebAssign (which I hate!) won't accept any form of that answer. Can someone set me straight?

Sketch the region in the xy-plane defined by the inequalities x − 3y² ≥ 0, 2 − x − 5|y| ≥ 0
and find its area.

 

[HallsofIvy]

The problem is copied and pasted below.

I think the area I am trying to find is bound by x=3y^2 on the left, x=2-5y on the right,

No. The fact that there is a an absolute value of y does NOT mean y is not negative.
If y> 0 the equation of the boundary is 2- x- 5y= 0 which is a straight line through (0, 2/5) and (2, 0), ending at (2, 0). If y< 0, the equation is 2- x- 5(-y)= 2- x+ 5y= 0 which is a straight line through (0, -2/5), (2, 0), again ending at (2, 0), connecting to the other line.

and the x axis. I then integrate from 0 to 1/3 (2-5y-3y^2)dy, and I get the answer 19/54. It seems so simple to me, but WebAssign (which I hate!) won't accept any form of that answer. Can someone set me straight?


Sketch the region in the

xy-plane defined by the inequalities x − 3y² ≥ 0, 2 − x − 5|y| ≥ 0
and find its area.

 

[AmySaunders]

So, then, I'm looking for that whole little piece's area, not just what is above the x axis? I guess I can see that. Sometimes it's hard to figure out what they want you to integrate. The integral is the easy part!

Thank you for your help!