Simplifying the Integral of a Piece of Helix Using Stokes' Theorem

[ashina14]

Homework Statement

Evaluate the integral
∫
(x2 − yz) dx + (y2 − xz) dy + (z2 − xy) dz, C(A→B)
where C(A → B) is a piece of the helix
x = a cos φ, y = a sin φ, z = h φ, (0 ≤ φ ≤ 2π),
2π
connecting the points A(a, 0, 0) and B(a, 0, h).

Homework Equations

[Hint: The problem could be tackled in different ways and, for one of them, Stokes’ theorem might be of some relevance.]

The Attempt at a Solution

acosφ =a, asinφ=0 therefore after subbing in for x, y and z I get:

Integral= ∫ a4 dx -(a2h φ)/2pi dy + (h2 φ2)/ 4 pi2 dz

But not sure what to do next

 

[jackarms]

Not sure what you mean by ##acos\phi = a## and ##asin\phi = 0## -- those are just for ##\phi = 0##. After you substitute in for x, y, and z, you also want to substitute dx, dy, and dz in terms of ##d\phi##

 

[HallsofIvy]

Homework Statement

Evaluate the integral
∫
(x2 − yz) dx + (y2 − xz) dy + (z2 − xy) dz, C(A→B)
where C(A → B) is a piece of the helix
x = a cos φ, y = a sin φ, z = h φ, (0 ≤ φ ≤ 2π),
2π
connecting the points A(a, 0, 0) and B(a, 0, h).

Your very first problem is that B is not on this helix!
Of course, taking [itex]\phi= 0[/itex] gives [itex](a cos(0), a sin(0), h(0)= (a, 0, 0)= A[/itex]
But in order to have [itex]z= h\phi= h[/itex], [itex]\phi[/itex] must be 1. But then [itex]x= a cos(1)[/itex] is NOT a and [itex]y= a sin(1)[/itex] is NOT 0!

Homework Equations

[Hint: The problem could be tackled in different ways and, for one of them, Stokes’ theorem might be of some relevance.]

The Attempt at a Solution

acosφ =a, asinφ=0 therefore after subbing in for x, y and z I get:

Integral= ∫ a4 dx -(a2h φ)/2pi dy + (h2 φ2)/ 4 pi2 dz

But not sure what to do next

Homework Statement

Homework Equations

The Attempt at a Solution

 

[ashina14]

Not sure how to go from there either, I am supposed to get a 0 in the end but am failing to do so. How could I do it with Stoke's thm instead?

 

[LCKurtz]

This doesn't look like a Stoke's theorem problem. But you likely have a misprint in your parameterization$$
x = a \cos \phi, y = a \sin \phi, z = h\phi$$If you change the last equation to ##z = \frac h {2\pi}\phi##, that will make the point ##B## be on the curve at ##\phi = 2\pi##.

 

[ashina14]

yes apparently it's a misprint! indeed z = hϕ/2π. I don't know the significance of B being the curve at ϕ=2π though.

 

[LCKurtz]

yes apparently it's a misprint! indeed z = hϕ/2π. I don't know the significance of B being the curve at ϕ=2π though.

If you are going to integrate between two points on the curve, then they need to be on the curve, don't they?

Your problem has nothing to do with Stoke's theorem. As a hint at what to do, have you checked the curl ##\nabla\times\vec F##, where ##\vec F## is the vector in ##\int_c\vec F\cdot d\vec R~##?

 

[ashina14]

Yes but I still don't feel the need to utilize the given points.

Here's a solution I've come up with

Changing the integrand in terms of
the angle @ through x=acos@; dx=-asin@d@, y=asin@;
dy=acos@d@, z=h@/(2pi); dz=hd@/(2pi), get

I [from @=0 to @=2pi]S[(a^2)cos^2@-ah@sin@/(2pi)]
(-asin@d@)+[(a^2)sin^2@-ah@cos@/(2pi)](...
+[(h@/(2pi))^2-(a^2)sin@cos@][hd@/(2pi)...

I=[from @=0 to @=2pi]S{[(a^3)(sin^2@cos@-cos^2@sin@)
+(a^2)h/(2pi)(1-2cos^2@)+[(a^2)h/(2pi)(...
d@=>

I={from @=0 to 2pi][((a^3)/3)(sin^3@)-cos^3@)=>
I=0.

Unless I've missed out on something, I don't think I needed the points/curl

 

[ashina14]

Although I see how it can be done with ∇×F⃗ , thanks!

 

[LCKurtz]

Yes but I still don't feel the need to utilize the given points.

Of course you have to use the two points. If the values of ##\phi## of ##0## and ##2\pi## don't give the two points, you are working the wrong problem.

Here's a solution I've come up with

Changing the integrand in terms of
the angle @ through x=acos@; dx=-asin@d@, y=asin@;
dy=acos@d@, z=h@/(2pi); dz=hd@/(2pi), get

I [from @=0 to @=2pi]S[(a^2)cos^2@-ah@sin@/(2pi)]
(-asin@d@)+[(a^2)sin^2@-ah@cos@/(2pi)](...
+[(h@/(2pi))^2-(a^2)sin@cos@][hd@/(2pi)...

I=[from @=0 to @=2pi]S{[(a^3)(sin^2@cos@-cos^2@sin@)
+(a^2)h/(2pi)(1-2cos^2@)+[(a^2)h/(2pi)(...
d@=>

I={from @=0 to 2pi][((a^3)/3)(sin^3@)-cos^3@)=>
I=0.

Unless I've missed out on something, I don't think I needed the points/curl

Trying to read that gives me a headache. All I can tell you is that zero is not the correct answer. And if you would think about my hint there is a much easier way.

 

[ashina14]

Yes they give the same two points. I don't need to explicitly use the points I feel though. Sorry for giving you a headache! Doing it your way is much shorter indeed but it's still giving me an answer of zero:

Taking vector F = (x² -yz) i + (y² -xz) j + (z² -xy)

so, taking cross product,
▽ x F = l i j k l
l P_x P_y P_zl
lx² -yz y² -xz z² -xyl

i(0) -j(0) +k(0) = 0

=> value of integral = 0

 

[ashina14]

|...i...j...k...l
|...Px...Py...Pz...|
|x^2-yz..y^2-xz..z^2-xy|

better version of product

 

[LCKurtz]

Yes they give the same two points. I don't need to explicitly use the points I feel though. Sorry for giving you a headache! Doing it your way is much shorter indeed but it's still giving me an answer of zero:

Taking vector F = (x² -yz) i + (y² -xz) j + (z² -xy)

so, taking cross product,
▽ x F = l i j k l
l P_x P_y P_zl
lx² -yz y² -xz z² -xyl

i(0) -j(0) +k(0) = 0

=> value of integral = 0

|...i...j...k...l
|...Px...Py...Pz...|
|x^2-yz..y^2-xz..z^2-xy|

better version of product

The curl of ##\vec F## being zero does not mean ##\int_C \vec F\cdot d\vec R=0## along every path ##C##. But it does imply some things that can simplify your problem, not to mention help you get the correct answer. What does your book tell you about this situation?