Double Integral - Volume question

[Sebs0r]

Homework Statement

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

Homework Equations

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

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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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

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

Yes, this is the correct integral

$$= \int ^{1}_{0} 4x - 4x^2 dx$$

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!

$$= 2/3$$

 

[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