# Multiple integral with substitution

1. Nov 21, 2009

### mlarson9000

1. The problem statement, all variables and given/known data
This is going to be confusing to read, as I don't know how to make this look right. The first integral is from 0 to L-2d, the second from x1+d to L-d, and the third from x2 to L. (F(x)=1)

1.) 0$$\int$$L-2d,x1+d$$\int$$L-d,x2+d$$\int$$L dx3dx2dx1

2.) 0$$\int$$L-2d,x1+d$$\int$$L-d (L-x2-d)dx2dx1

3.) substituting y2 for L-x2-d

4.) 0$$\int$$L-2d,0$$\int$$L-X1-2d y2dy2dx1

2. Relevant equations

3. The attempt at a solution
The answer to this isn't that important, as I already have the solution. What I don't understand is why after substituting y2, the upper and lower bounds of the corresponding integral aren't the other way around, with L-x1-2d the lower bound, and 0 the upper. Is this because dy2=-dx2? Probably, but I just want to make sure I understand what's going on.

Last edited: Nov 21, 2009
2. Nov 22, 2009

### HallsofIvy

Staff Emeritus
So this is
$$\int_0^{L-2d}\int_{x_1+d}^{L-d}\int_{x_2+d}^L dx1dx2dx3$$?
(You say above "the third from x2 to L" but have "x2+ d" in the integral)
Click on the LaTex to see the code.

Okay, good.

If y2= L- x2- d, then when x2= L-d, the upper limit, y2= L-(L-d)-d= L-L+d-d= 0. When x2= x1+ d, the lower limit, y2= L- (x1+d)-d= L-x1-d-d= L- x1- 2d.

However, just as you say, dy2= -dx2 and :
$$\int_a^b f(x)dx= -\int_b^a f(x)dx$$

Taking that "-" out reverses the limits of integration.

3. Nov 22, 2009

### tiny-tim

eugh! :yuck:

Either use the X2 and X2 tags just above the Reply box …

0L-2d

or type \int_{}^{} in LaTeX …

$$\int_{0}^{L-2d}$$

Hi mlarson9000!
Yes, it's because dy2 = -dx2

but I personally think that's badly and confusingly written , and the author should have kept the limits the same way up, and put a minus in front of the integral.