How Do You Calculate the Surface Area of a Helix?

[Legolaz]

Hi Forum,
Just wanted to ask how to compute surface area of a helix x =cos t, y =sin t, z=t by integration
Thanks in advance.

 

[RUber]

I think what you described sounds more like a contour. Normally, you would find the length of that. Is there some way you can define the surface?

 

[Ssnow]

yeah, you need another way to define the surface, also in this case you must specify the range of $t$, I suppose it is finite.

 

[RUber]

If you are looking for the contour length, that follows the simple formula:
$\int_{t_0}^{t_{final}} \sqrt{ \left( \frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2 } \, dt$
Which for initial time = 0, and final time = t, should give you a contour length of $L = t\sqrt{2}$

 

[Legolaz]

Thank you Ruber.
I already figured out Area would be just = ∫∫ rdr dθ

 

[Mark44]

Thank you Ruber.
I already figured out Area would be just = ∫∫ rdr dθ

This doesn't make any sense. A helix is essentially a one-dimensional curve in three-dimensional space. A helix doesn't have "area".

 

[Legolaz]

Yes it does. See figure left, where the ball slides on the surface.

 

[Mark44]

A helix doesn't rotate, unlike what is shown in your animation. The helix in your OP, x = cos(t), y = sin(t), z = t, just sits there.

 

[Legolaz]

Stay focus on the topic about helix surface area. The rotation is just a representation of the area. Equation(x=cost, y=sint, z=t) is the ideal equation for a helix.

 

[RUber]

There are a few things to consider here.
1) In the graphic you posted, there was a central pole with radius R1, which I assume would not contribute to the area you are interested in. The outer radius of your "ideal" equation is 1, since the plot traces the unit circle. The shape you are interested is the surface traced from (R1cos t, R1sin t , z ) to (cos t, sin t, t) as t goes from A to B.
2) You still have not described the total height you are interested.
3) If you consider the interior radius R1 to be zero, then you are tracing a unit circle every 2pi units in t. So...the area of the helix is T/2pi where T = B-A, the total interval length in T. Of course, you could scale this using the simple area of a circle formula for different radii.

 

[Legolaz]

Yes, Ruber, I got it. Thank you for the inputs.

 

[Mark44]

Stay focus on the topic about helix surface area.

I am focussed on the topic. I understand what you're trying to find, but the problem shouldl have been stated differently, maybe something like this: "A helical screw whose edge is described by x = cos(t), y = sin(t), and z = t, rotates about its central axis. Find the area of the cylinder swept by this screw."

Additional information about the length of the screw is needed, as RUber points out.

The rotation is just a representation of the area. Equation(x=cost, y=sint, z=t) is the ideal equation for a helix.

 

[Legolaz]

Yes, i apologize for the lacking info Mark44 and do understand my english, simply because its not my native tongue, it sounds rude and inappropriate sometimes though.