Area of a Sinus Curve using Pythagoras trigonometry

Question

Good day.

I have devised a method for obtaining the Area of a Sinus curve using Pythagoras trigonometric formulas. Are anyone familiar with this method. Take a look.

By transforming the sinus into a paper cylinder, I can calculate the circumference of the cylinder by using the Radius, that is equal to the Amplitude. The wavlenght of one period is therefore the lenght of the cylinder. Making a paper model I see that the lenght of the Sinus curce is equal to the Diameter of the cylinder rolled up into a plane square area.

$$f = 3 \sin ( 4\pi x )$$

Anatomy.

$3 \sin$ = Amplitude, Radius

$x$ = wave is along the $x$ axis

$x\pi = 1$ half wave is $1$ grid unit long

$4$ = four half waves per grid unit.

I can with one line of code, calculate the area of half a wave, the entire period, or any lenght of the North and South polarisations .

$$f = 3 \sin ( 4\pi x )$$ Radius = $3$ cm

Limit_end = I want 1 period = $\frac{1}{4} \cdot 2 = 0.5$ cm

Formula Area Curve = Lim * 2PiR / Pi^2

Area Curve = $\frac{(0.5)(2)(\pi)(3)}{\pi^2}$

Area Curve = $\frac{(0.5)(2)(3)}{\pi} = 0.954 cm^2$ .

X_lim_start = 0 . X_lim_end = 0.5

Im not too familiar with calculus. And This method described I presume only works for Sinus, and functions with Radius. Not oddly shaped geometries or other functions .

Answer 1

Using calculus, we see that the area under half a period of the sinusoidal function $f(x) = a \sin \omega x$ is $$ \int_0^{\pi/\omega}a \sin(\omega x)\,dx = \frac{2a}{\omega} $$ This agrees with your calculations.

But I'm having a hard time understanding this paragraph:

By transforming the sinus into a paper cylinder, I can calculate the circumference of the cylinder by using the Radius, that is equal to the Amplitude. The wavelength of one period is therefore the length of the cylinder. Making a paper model I see that the length of the Sinus curve is equal to the Diameter of the cylinder rolled up into a plane square area.

Working backwards, it sounds like you're deriving a formula that the area under a half-period is equal to $\frac{2\pi a}{\omega \pi}$. But without more details about how you “transform” the sine curve into a cylinder, it's hard to verify that process.

Answer 2

Thanks you for the reply. I will explain what I mean, by "transforming the sinus into a cylinder" .

A while ago, I saw that if I had a square drawn on a paper, then drew in the diagonal. I clipped out the square, then rolled the square into a cylinder. I saw that the diagonal became a Sinus shape, on the paper cylinder. The 2 dimensional square was "transformed" into a 3 dimensional cylinder. and a straight line diagonal "transformed" into a sinus wave. I was amazed .

And I thought there was a relationship between the Diagonal of a square and the Sinus . 2 years went by and a few days ago I started again on the problem. And by sheer luck (!) I divided the the Circumference of that cylinder with the "wavlenght" or lenght of one period, by Pi^2 ! And the Area of the Sinus curve matched the result from a Caculus approach . I was also amazed that the Radius of the sinewave would be the same as signal amplitude .

Another way of looking at it, If I have a cylinder, made of paper. Then photocopy a sinus onto that cylinder. Then fold the cylinder out into a plane square. The Sinus becomes a Diagonal, or Hypothenus . If the Sinus wavepeak = 1 on that "cylinder", unwrapping it into a square, makes the =>

Area_square in cm or inches : X axis 6.28 cm * Y axis 6.28 cm.

Paper square area = 2Pi cm * 2Pi cm.

Holding a ruler close to the cylinder will measure sine wavepeak = 1 cm , due to the curvature of pi. Its a geometric approach . And fun for young people, where you have to clip ut a paper and see the measures come to life . Like a practical experiment . Or for people who needs a practical approach to understaning things, like myself .

To me it was a revelation, and maybe a way for people to learn calculus or desire to learn electricity and phases . Concerning Rotors and RPM and hertz, amplitudes and wavelenght . On my profile there is a link to my blog on blogspot . the blog link in the post was administered away .

Answer 3

If you study the following five relations carefully I believe it has answers to first part of your question.

Sine curve equation in developed form

$$ z= A \sin \frac{2 \pi x }{\lambda} \tag1 $$

Basis of circumferential wavelength

$$ \lambda = 2 \pi R \tag2 $$

plug the above into(1)

Relation of polar angle and circumferential length

$$ \frac {x}{R} = \theta \tag3 $$

plug the above into (1)

Spatial equation of a helix is obtained as:

$$ z= A \sin \theta, x = R \cos \theta,\, y = R \sin \theta \tag4 $$

Eliminate $R$ from first and last of above three equations

$$ \frac{z}{y} = \frac{A}{R} = \tan \alpha =\frac{z_{max}}{y_{max}} = \frac{h}{R} \tag5 $$

So a sine curve is produced as curve of intersection by cutting a cylinder with a plane inclined at $\alpha $ to diametrical plane of the basis cylinder and laying it flat.

The second part, your area computation I cannot follow. You took $ \pi=3 $ or something..?

Answer 4

It looks very much like it is the same thing. I am interpreting this artistic way .

Drawing of Area of Sinus Curve, with respect to baseline X-Axis

When the "cylinder" is unwrapped. the squares lenght and height, X-axis and Y-axis is multiplied then divided by pi^2. Area sin = X * Y / Pi^2

A cylinder of clay, is rolled over a square with an embossed diagonal. A sine curve will be imprinted on the clay roll . I cant explain it any better than to say what I did, to obtain the method .

Pardon my facination for herioglyphs and all that, its a gratitude to Ptha, the egyptian god of engeneering, and Ra, that I regard is the origin to Radius, or the Sun . And it says supposedly "sinus" in glyphs.