Integrating to an absolute value?

In summary, the integral is trying to find the electric field at a point on the axis of a rod when the polarization is known, but the problem is that the integrand does not seem to be giving the correct answer.
  • #1
Blastrix91
25
0

Homework Statement


I need some help understanding an integral step in the example below. I get how the integrand was set up, but I don't get how comes to two expressions with the absolute value of z-L/2.

(Problem description if that is needed: L is the length of a cylinder with radius R, and P is the polarization of the cylinder in direction of its length. Calculate the electric field at a point on the axis of the rod)

Homework Equations



http://imageshack.us/a/img594/1895/absolutevalue.png

The Attempt at a Solution


I don't understand that step really. Google has no answers. I thought that it might have been some absolute value identities, but it seems that was not the case. Is there someone who'd lend a hand?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Blastrix91 said:

Homework Statement


I need some help understanding an integral step in the example below. I get how the integrand was set up, but I don't get how comes to two expressions with the absolute value of z-L/2.

(Problem description if that is needed: L is the length of a cylinder with radius R, and P is the polarization of the cylinder in direction of its length. Calculate the electric field at a point on the axis of the rod)

Homework Equations



http://imageshack.us/a/img594/1895/absolutevalue.png

The Attempt at a Solution


I don't understand that step really. Google has no answers. I thought that it might have been some absolute value identities, but it seems that was not the case. Is there someone who'd lend a hand?

[tex] \int_0^R \frac{r}{\sqrt{r^2+a^2}} dr
= \frac{1}{2} \int_{r=0}^R \frac{d(r^2)}{\sqrt{r^2+a^2}}
= \frac{1}{2} \int_0^{R^2} \frac{dy}{\sqrt{y+a^2}} = \left. \sqrt{y+a^2}\right|_0^{R^2}[/tex] Look at what you get if a > 0 or a < 0.

RGV
 
Last edited by a moderator:
  • #3
Blastrix91 said:

Homework Statement


I need some help understanding an integral step in the example below. I get how the integrand was set up, but I don't get how comes to two expressions with the absolute value of z-L/2.

(Problem description if that is needed: L is the length of a cylinder with radius R, and P is the polarization of the cylinder in direction of its length. Calculate the electric field at a point on the axis of the rod)

Homework Equations



http://imageshack.us/a/img594/1895/absolutevalue.png

The Attempt at a Solution


I don't understand that step really. Google has no answers. I thought that it might have been some absolute value identities, but it seems that was not the case. Is there someone who'd lend a hand?

It's because, [itex]\displaystyle \sqrt{u^2}=|u|\ .[/itex]
 
Last edited by a moderator:
  • #4
Wow thank you. That was pretty helpful
 

1. What does it mean to integrate to an absolute value?

Integrating to an absolute value means finding the area under a curve but taking into account the direction of the curve. This results in a positive value, regardless of the direction the curve is moving.

2. Why is it important to integrate to an absolute value?

Integrating to an absolute value is important because it allows for a more accurate calculation of the area under a curve. Without taking into account the direction of the curve, the calculated area could be negative, which would not accurately represent the actual area.

3. How is integrating to an absolute value different from regular integration?

Regular integration finds the area under a curve without considering the direction of the curve. Integrating to an absolute value takes into account the direction of the curve, resulting in a positive value.

4. What types of functions require integrating to an absolute value?

Functions that involve absolute values, such as |x| or |sin(x)|, require integrating to an absolute value. These functions have both positive and negative values, so integrating without considering the direction of the curve would result in an inaccurate calculation of the area.

5. How is integrating to an absolute value used in real-world applications?

Integrating to an absolute value is commonly used in physics and engineering to calculate displacement, velocity, and acceleration. It is also used in economics to calculate net benefit or loss. Essentially, any situation where direction matters and a positive value is needed, integrating to an absolute value is necessary.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
961
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
708
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
13K
Back
Top