Circular Helix Line Integral: Solving with r and dr/dλ

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
ferret123
Messages
23
Reaction score
0

Homework Statement



don't know the line integral latex code but;

[itex]\int[/itex][itex]\underline{r}[/itex][itex]\times[/itex]d[itex]\underline{r}[/itex]

from (a,0,0) to (a,0,2∏b) on the circular helix [itex]\underline{r}[/itex] = (acos(λ), asin(λ), bλ)

The Attempt at a Solution



Its the multiple use of the position vector r in the question that's confusing me. So far I've tried paramaterising the original integral as (r cross dr/dλ)dλ with dr/dλ being the derivative of the circular helix however I am confused as to whether the r in the integral is the same as the one describing the helix.

Am I on the right track or will i need to use another method?
 
Physics news on Phys.org
ferret123 said:

Homework Statement



don't know the line integral latex code but;

[itex]\int[/itex][itex]\underline{r}[/itex][itex]\times[/itex]d[itex]\underline{r}[/itex]

from (a,0,0) to (a,0,2∏b) on the circular helix [itex]\underline{r}[/itex] = (acos(λ), asin(λ), bλ)

The Attempt at a Solution



Its the multiple use of the position vector r in the question that's confusing me. So far I've tried paramaterising the original integral as (r cross dr/dλ)dλ with dr/dλ being the derivative of the circular helix however I am confused as to whether the r in the integral is the same as the one describing the helix.

Am I on the right track or will i need to use another method?

The r given to you is the parametric representation of the helix. It is easy to check that this is a suitable parametrisation.