Can someone just check if I did this line integral correctly?

Click For Summary
SUMMARY

The discussion centers on the evaluation of the line integral $$\int _{ 0 }^{ 2\pi }{ xdx }$$, where the variable x is defined as $$x = \cos(t)$$ and $$dx = -\sin(t) dt$$. The integral evaluates to zero due to the periodic nature of sine and cosine functions over the interval from 0 to 2π, resulting in the cancellation of positive and negative areas. It is clarified that this integral represents the integration of x around a circle, not the integral of x itself over the same interval, which would yield a different result.

PREREQUISITES
  • Understanding of line integrals in calculus
  • Familiarity with parametric equations and their derivatives
  • Knowledge of periodic functions, specifically sine and cosine
  • Basic concepts of area under curves and cancellation in integrals
NEXT STEPS
  • Study the properties of line integrals in vector calculus
  • Learn about parametric equations and their applications in integration
  • Explore the concept of periodic functions and their integrals
  • Investigate the geometric interpretation of integrals over closed curves
USEFUL FOR

Students studying calculus, particularly those focusing on line integrals and parametric equations, as well as educators seeking to clarify common misconceptions in integral calculus.

ainster31
Messages
158
Reaction score
1

Homework Statement



S4Fimyg.png


Homework Equations





The Attempt at a Solution



$$\int _{ 0 }^{ 2\pi }{ xdx } \\ =-\int _{ 0 }^{ 2\pi }{ sintcostdt } \\ =0$$

It feels wrong.
 
Physics news on Phys.org
I got zero also using a different method. Let's review the steps. You said
x = cos t, dx = -sint dt and t now runs from 0 to 2##\pi##.

Since cost and sint are both periodic in 2##\pi## you got zero.

What may feel wrong is that ##\int_0^{2\pi}x dx## is not zero. However, this is not the same thing as integrating x around a circle. Basically you are running cos(t) around that circle, and it's going to start and end with the same value.
 
Notice that x is positive on the right side of the circle, negative on the left side of the circle. The two halves cancel, leaving 0.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K