Multivariable Calculus, Line Integral

Click For Summary
SUMMARY

The discussion focuses on calculating the line integral of the vector field $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$ along the path $$C$$ defined by $$\dfrac{(\cos(t), \sin(t))}{1 + e^t}$$ for $$0 ≤ t < \infty$$. Participants emphasize the importance of substituting the parameterized path into the integral and breaking down the vector field into its x and y components. The final integral to evaluate is $$\int_0^{\infty} (F_x(x(t),y(t))x'(t)+F_y(x(t),y(t))y'(t))dt$$, which may simplify through trigonometric identities.

PREREQUISITES
  • Understanding of vector fields and line integrals
  • Familiarity with parameterization of curves
  • Knowledge of differentiation and trigonometric identities
  • Basic skills in multivariable calculus
NEXT STEPS
  • Practice evaluating line integrals in vector fields
  • Study parameterization techniques for curves in multivariable calculus
  • Explore trigonometric identities and their applications in calculus
  • Learn about the Fundamental Theorem for line integrals
USEFUL FOR

Students and educators in multivariable calculus, mathematicians working with vector fields, and anyone seeking to deepen their understanding of line integrals and their applications.

tompenny
Messages
15
Reaction score
3
Homework Statement
Calculate a line integral
Relevant Equations
$$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$
The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
tompenny said:
Homework Statement:: Calculate a line integral
Relevant Equations:: $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
Remember, ##\int_C \vec F \cdot d\vec r = \int_t \vec F(\vec r(t))\cdot \vec r'(t)~dt##. So the obvious hint is plug the formulas in and see what happens. Then come back to show us where you are stuck. There is no substitute for getting your hands dirty.
 
  • Like
Likes   Reactions: Delta2 and etotheipi
My advice would be to split ##\mathbf{F}## as well the path ##C## into x and y components. For example it will be $$F_x(x,y)=\frac{x}{\sqrt{1-x^2-y^2}}$$ $$F_y(x,y)=\frac{y}{\sqrt{1-x^2-y^2}}$$while for the path C it will be $$x(t)=\frac{\cos t}{1+e^t}$$ $$y(t)=\frac{\sin t}{1+e^t}$$. Then calculate $$x'(t)$$ and $$y'(t)$$ (derivatives of x(t),y(t)with respect to t). Then combine the whole thing to $$\int_0^{\infty} (F_x(x(t),y(t))x'(t)+F_y(x(t),y(t))y'(t))dt$$.
This might look a bit scary but i think if you do all the calculations correctly and the algebra correctly some simplifications will occur (by the use of trigonometric identities).
 
  • Love
Likes   Reactions: etotheipi

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
1K
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
6
Views
3K