Line Integral Problems on Ellipse Boundary | Compute Integrals Directly

Click For Summary
SUMMARY

The discussion focuses on computing three line integrals directly around the boundary of an ellipse defined by the equation (x^2/a^2)+(y^2/b^2)=1 in the first quadrant. The integrals computed are: (a) integral(xdy-ydx) resulting in ab(pi), (b) integral((x^2)dy) yielding 0, and (c) integral((y^2)dx) equating to -(4/3)a(b^2). The user questions whether the bounds for these integrals should be from 0 to pi/2 instead of 0 to pi, as they are only considering the first quadrant.

PREREQUISITES
  • Understanding of line integrals in vector calculus
  • Familiarity with parametric equations and their applications
  • Knowledge of the ellipse equation and its properties
  • Basic integration techniques in calculus
NEXT STEPS
  • Study the application of Green's Theorem in line integrals
  • Learn about parametrization techniques for different curves
  • Explore the properties of integrals over closed curves
  • Investigate the implications of changing integration bounds in parametric integrals
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and vector analysis, as well as anyone involved in solving line integral problems in physics or engineering contexts.

JaysFan31

Homework Statement


This is my problem:
Compute the following three line integrals directly around the boundary C of the part R of the interior ellipse (x^2/a^2)+(y^2/b^2)=1 where a>0 and b>0 that lies in the first quadrant:
(a) integral(xdy-ydx)
(b) integral((x^2)dy)
(c) integral((y^2)dx)


Homework Equations


I used parametrisation (x=acost and y=bsint) for the arc of the ellipse.
C is the curve r=(acost)i + (bsint)j (0 less than or equal to t less than or equal to pi).


The Attempt at a Solution


(a) integral(xdy-ydx)=integral from 0 to pi((acost)(bcost)dt)- integral from 0 to pi((bsint)(-asint)dt)=(ab)pi/2-(-ab)(pi)/2=ab(pi)

(b) integral((x^2)dy)=integral from 0 to pi((acost)(acost)(bcost)dt)=0.

(c) integral((y^2)dx)=integral from 0 to pi((bsint)(bsint)(-asint)dt)=-(4/3)a(b^2)

Could anyone check these and see if they are right?
 
Physics news on Phys.org
Actually, should all the bounds be from 0 to pi/2 instead of 0 to pi (since I am looking at only the first quadrant)?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
12
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
2K