Simple Line Integral becomes troublesome

Click For Summary

Homework Help Overview

The discussion revolves around evaluating a line integral involving a parametrized curve defined by x=9cos(t), y=9sin(t), and z=-8t, with t ranging from 0 to 10π. The integral to be calculated is ∫_C (7x² + 4y² - 5xy) ds, which has led to some confusion regarding the correct setup and evaluation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to transform the integral based on the parametrization but expresses concern over the complexity of the resulting expression. Other participants suggest checking boundary conditions for t and clarify the need for the differential in the integral, questioning whether it should be with respect to dx, dy, dz, or arclength ds.

Discussion Status

Participants are actively engaging with the problem, with one noting a mistake in their calculations related to the differential. There is a recognition of the need to correctly apply the Pythagorean identity in the context of the integral, which has been identified as a key issue affecting the original poster's progress.

Contextual Notes

There is an emphasis on ensuring the correct interpretation of the differential in the integral, as well as the implications of the parametrization on the evaluation process. The discussion reflects a collaborative effort to clarify misunderstandings and refine the approach to the problem.

gipc
Messages
69
Reaction score
0
I have the parametrization of C
x=9cos(t)
y=9sin(t)
z=-8t

0<=t<=10*pi

and I have to calculate \int_c 7x^2+4y^2 -5xy

after I transform this I get
(7*81cos^2(t) +4*81sin^2(t) -5*81*cos(t)*sin(t))*sqrt(-9sin^2(t)+9cos^2(t)-8) dt from 0 to 10*pi

Now that is a monster.
What did I do wrong?
 
Physics news on Phys.org
Obtain the values of x, y, z in terms of the boundary conditions for t, i.e. t = 0 and t = 10\pi. It should be easier to solve.
 
When t= 0 x= 9, y= 0, z= 0.

When t= 10\pi, x= 9, y= 0, z= -80\pi.

I don't see how that helps at all!

gipc, you statement \int_c 7x^2+ 4y^2- 5z^2 is missing the differential. Is that with respect to dx, dy, dz, or the arclength ds, or what?

Because of the square root, I assume it is ds but it would not be what you give. it would be
\sqrt{81 sin^2(t)+ 81 cos^2(t)+ 64}dt= \sqrt{81+ 64}dt= \sqrt{145}dt

Did you forget that sin^2(t)+ cos^2(t)= 1?
 
yes, i did forget :)
thanks for that, that was the mistake that was holding me back
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K