# Can I change the limits of this double integral

1. Jan 8, 2013

### egroeg93

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

R is the region bounded by y=x^2 and y=4. evaluate the double integral of f(x,y)=6x^2+2y over R

After drawing the region I was wondering if I could just work with the first quadrant and then double my solution, because both y=x^2 and y=4 are even functions so my question is does my solution work? If so would my very first line be correct? Oh and I'm not sure how to write what I'm integrating between so when I put ∫[a,b]f'(x)dx thats f(a)-f(b).

2. Relevant equations

3. The attempt at a solution

∫[4,0]∫[y^1/2,-y^1/2]6x^2+2y dxdy = 2*∫[4,0]∫[y^1/2,0] 6x^2+2y dxdy
following it through
= 2*∫[4,0] [[2(y^1/2)^3+2*y*y^1/2]-[0]] dy
= 2*∫[4,0] 4*y^3/2 dy
= 2*[[8/5*(4^5/2)]-[0]]
= 512/5

2. Jan 8, 2013

### lurflurf

That looks right at a glance. You can split it, but it does not help much.

write

$$\int_{-2}^2 \int_{x^2}^4 \! (6x^2+2y) \text{ dy dx}$$

or just

$$\int \limits_A \! (6x^2+2y) \text{ dA}$$

if you can't be bothered with the limits.

3. Jan 8, 2013

### egroeg93

Okay let me rephrase my question with the aid of proper symbols :)

$$\int_{-2}^2 \int_{x^2}^4 \! (6x^2+2y) \text{ dy dx}$$ =
$$2*{\int_{0}^4 \int_{0}^{y^{1/2}} \! (6x^2+2y) \text{ dx dy}}$$

Is this just coincidence or is this always true when you have a Region formed by two even functions as in the question?

4. Jan 8, 2013

### lurflurf

Yes that is true (with some conditions) it is called Fubini's theorem. It is not quite always true if you include improper integrals, for example this common example given in the above Wikipedia link.

$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{ dx dy}=-\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{ dy dx}$$

That is the type of thing considered in theoretical treatments of calculus, but it is good to be aware of. When the function either does not go to infinity or is absolutely integrable it is safe to interchange the integrals. Of course (6x^2+2y) is a very well behaved integrant. What you have done is express the region in two equivalent ways.

2<x<-2 and x^2<y<4
is the same as
0<y<4 and -sqrt(y)<x<sqrt(y)

and your symmetry argument that the integral over half the region equals half the integral over the whole region.

Last edited: Jan 8, 2013