- #1

- 32

- 3

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.

- Thread starter Legolaz
- Start date

- #1

- 32

- 3

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.

- #2

RUber

Homework Helper

- 1,687

- 344

- #3

Ssnow

Gold Member

- 523

- 154

- #4

RUber

Homework Helper

- 1,687

- 344

## \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}##

- #5

- 32

- 3

Thank you Ruber.

I already figured out Area would be just = ∫∫ rdr dθ

I already figured out Area would be just = ∫∫ rdr dθ

- #6

Mark44

Mentor

- 33,982

- 5,640

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

I already figured out Area would be just = ∫∫ rdr dθ

- #7

- 32

- 3

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

Last edited by a moderator:

- #8

Mark44

Mentor

- 33,982

- 5,640

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.

Last edited by a moderator:

- #9

- 32

- 3

- #10

RUber

Homework Helper

- 1,687

- 344

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.

- #11

- 32

- 3

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

- #12

Mark44

Mentor

- 33,982

- 5,640

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."Stay focus on the topic about helix surface area.

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

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

- #13

- 32

- 3

- Last Post

- Replies
- 1

- Views
- 11K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 3

- Views
- 549

- Replies
- 23

- Views
- 12K

- Replies
- 4

- Views
- 766

- Last Post

- Replies
- 2

- Views
- 568

- Last Post

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 7K

- Last Post

- Replies
- 12

- Views
- 4K

- Last Post

- Replies
- 2

- Views
- 1K