Area between curves

Question

Sketch the region in the xy-plane defined by the inequalities
x − 3y2 ≥ 0, 2 − x − 5|y| ≥ 0
and find its area.
If I did it correctly, I have a region defined by x=3y^2, (x-2)/5, -(x-2)/5 with x values from 0-2 and y values from -1/3-1/3. 
Pi then divided the region from x=0-1/3 and 1/3-2. The first portion, I integrated with respect to y --3(y^2) from -1/3 to 1/3 and got 2/27.

Next, I thought I could integrate -(x-2)/5 with respect to x  from 1/3 to 2 and then double it to get the second area. I was getting a negative number so clearly that wasn't right. Then I tried integrating it with respect to y and doubling it and adding it to the first area. My answer is wrong, but I don't know where I went wrong. 

Could you track my basic steps and tell me if I even interpreted the graph correctly? And could you walk me through  The problem please?

Thanks

Answer 1

Rather than track your steps, I will show you how I would approach this problem. 

Areas can be computed as integrals over either x or y.    For the problem at hand, it is easier to think of x as function of y and to carry out the integral over y.  Notice that the equations do not change if y is replaced by -y.  This means that the area bounded by the curves has the x axis as an axis of symmetry.   If we plot x on the vertical axis and y on the horizontal axis, the vertical axis divides the area into a left half and a right half that are mirror images of each other with equal areas.  Taking advantage of this, I will compute the area of the right half and then double the answer as a final step.    For the right half side y is always positive, so the boundary equations can be written as

   (upper boundary)    x = 2 - 5y

   (lower boundary)    x = 3 y^2

The boundaries cross at   y = 1/3  , x = 1/3  and the vertical axis (x axis) forms another boundary for the right half side.   Thus the area of the right half is

 

 ∫ [ 2-5 y - 3 y^2]  dy     with limits   0 and 1/3

 

The integral is elementary   yielding

 

   [ 2 y - (5/2) y^2 - (3/3) y^3 ]       evaluated at 1/3  and 0.   The result is:

 

  area of right half  =  19/54.       So the whole area is twice this result

 

  Total area =   19/27