Multiple integral with substitution

In summary: But the author didn't do that, so we have to live with it.In summary, the author uses a substitution to change variables in the integral, which results in dy2 = -dx2, causing the upper and lower limits to switch positions in the corresponding integral. This is because the substitution affects the orientation of the integration, requiring a negative sign in front of the integral.
  • #1
mlarson9000
49
0

Homework Statement


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[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d,x2+d[tex]\int[/tex]L dx3dx2dx1


2.) 0[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d (L-x2-d)dx2dx1

3.) substituting y2 for L-x2-d

4.) 0[tex]\int[/tex]L-2d,0[tex]\int[/tex]L-X1-2d y2dy2dx1

Homework Equations





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.
 
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  • #2
mlarson9000 said:

Homework Statement


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[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d,x2+d[tex]\int[/tex]L dx3dx2dx1
So this is
[tex]\int_0^{L-2d}\int_{x_1+d}^{L-d}\int_{x_2+d}^L dx1dx2dx3[/tex]?
(You say above "the third from x2 to L" but have "x2+ d" in the integral)
Click on the LaTex to see the code.


2.) 0[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d (L-x2-d)dx2dx1.
Okay, good.

3.) substituting y2 for L-x2-d

4.) 0[tex]\int[/tex]L-2d,0[tex]\int[/tex]L-X1-2d y2dy2dx1

Homework Equations





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.
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 :
[tex]\int_a^b f(x)dx= -\int_b^a f(x)dx[/tex]

Taking that "-" out reverses the limits of integration.
 
  • #3
eugh!

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

0L-2d

or type \int_{}^{} in LaTeX …

[tex]\int_{0}^{L-2d}[/tex] :smile:

Hi mlarson9000! :smile:
mlarson9000 said:
… 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.

Yes, it's because dy2 = -dx2

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

Related to Multiple integral with substitution

1. What is multiple integral with substitution?

Multiple integral with substitution is a technique used to evaluate integrals of multivariable functions by substituting variables with other variables or functions. It is an extension of single variable integration and is commonly used in physics, engineering, and other scientific fields.

2. Why is multiple integral with substitution important?

Multiple integral with substitution is important because it allows us to solve complex integrals that cannot be solved using traditional integration techniques. It also has many applications in physics and engineering, such as calculating volumes, areas, and moments of inertia.

3. How does multiple integral with substitution work?

To use multiple integral with substitution, we first identify the variables to be substituted and then determine the appropriate substitution function. We then apply the substitution to the original integral and solve for the new integral, which can be evaluated using traditional integration techniques.

4. What are some common substitution functions used in multiple integral with substitution?

Some common substitution functions used in multiple integral with substitution include polar, cylindrical, and spherical coordinates. These coordinate systems are particularly useful when dealing with problems involving circles, cylinders, and spheres.

5. What are some tips for using multiple integral with substitution effectively?

To use multiple integral with substitution effectively, it is important to carefully choose the substitution function and to check the limits of integration. It is also helpful to visualize the problem in terms of the new coordinate system to better understand the integral. Practice and familiarity with different substitution techniques can also improve efficiency when solving multiple integrals.

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