# Double Integral Over General Region

1. Nov 4, 2009

### vandyboy73191

1. The problem statement, all variables and given/known data

1. Find the volume of the solid which is under the surface z = 2x + y2 and above the region bounded by x = y^2 and x = y^3.

2. Relevant equations

3. The attempt at a solution

So first I graphed x=y^3 and x=y^2. (http://h.imagehost.org/view/0716/Math_Problem [Broken])
I found their points of intersection (y=1 or y =0).
Set up double integral as Integral from 0 to 1 Integral from y^2 to y^3 of (2x+y^2) dx dy
where y^2<x<y^3 and 0<y<1

I calculated the integral and got 1/7 plus 1/6 minus 2/5

Is my work correct?

Last edited by a moderator: May 4, 2017
2. Nov 4, 2009

### Staff: Mentor

Your answer is almost right; your sign is wrong. It should be 2/5 - 1/7 - 1/6 = 19/210
For each horizontal strip, the left boundary is x = y^3 and the right boundary is x = y^2. You have them reversed in your inner integral, which gives you the opposite sign.

3. Nov 4, 2009

### union68

Based on your picture, shouldn't it be $$y=x^3$$ and $$x=y^2$$?

However, if you did write the equations correctly, then you've drawn the region wrong.

4. Nov 4, 2009

### Staff: Mentor

vandyboy's graph for x = y^3 is incorrect. He has actually drawn the graph of y = x^3.

5. Nov 4, 2009

### vandyboy73191

Oops. Yeah I just noticed that. thanks guys