# Double Integral - Volume question

1. Jun 4, 2009

### Sebs0r

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

Find the volume of the solid bounded by z = 0 and z = 2xy, lying in the first quadrant and bounded by the curves y = x^2 and x+y = 2

2. Relevant equations

3. The attempt at a solution
I have an answer, but just asking if I've done it correctly, since we arent given the solution:

Intersection of x^2 and 2-x -> x = 1 or x = -2

Limits
x^2 <= y <= 2-x
0 <= x <= 1
0 <= z <= 2xy

$$Volume = \int ^{1}_{0}\int^{2-x}_{x^2} (2xy - 0) dy dx$$
$$= \int ^{1}_{0} 4x - 4x^2 dx$$
$$= 2/3$$

Last edited by a moderator: Jun 4, 2009
2. Jun 4, 2009

### HallsofIvy

Staff Emeritus
Yes, this is the correct integral

How did you get this? The integral of y will involve y2 and letting y= x2 for the lower limit will give x4. Multiplying by the x in the original integral, this integrand should have an x5 in it!

3. Jun 4, 2009

### Sebs0r

Right -> I mistook the x^2 for an x :p. Trying to do too many steps in your head inevitably leads to errors.
So the new integral then is

$$\int^{1}_{0} x^3 - 4x^2 + 4x - x^5 dx$$

3/4

Should be right :p
Thanks man