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I had to integrate an area delimited by a quarter of a circle, something like this:

http://www.wolframalpha.com/input/?i=integrate+10+-+sqrt%2864+-+x%5E2%29+dx+from+0+to+5

Which comes from the equation:

$$x^2 + (y-10)^2 = 8^2$$

I wondered if it was possible to express the same curve but using sin instead?

I read somewhere (cannot find it again) you could use something like:

$$y = 10 - 8∗\cos(xπ/(2∗8))$$

But it seems just wrong? Can you confirm that it's not possible to do a perfectly circle-shaped form using sin/cos using "classic" equations (not parametric equations).

[EDIT]

I'm not interested in polar coordinates either... I want to know if it's possible to have a final equation in the form of:

$$y = A + B * \sin(C * x)$$ for some values of A/B/C. You are free to add some cos() in there :)

[EDIT2]

Apparently the question I should have asked is: "Is it possible to express parts of a circle as a cartesian equation involving sin/cos"

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    $\begingroup$ Substituting $x=8\sin\theta$ will lead you to the required antiderivative. $\endgroup$ – David Mitra Jan 21 '14 at 13:01
  • $\begingroup$ I'm sorry can you elaborate? wouldn't introducing an angle make it a parametric equation? $\endgroup$ – Silex Jan 21 '14 at 13:07
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Recall polar coordinate equivalents.

Here, radius is $r=8,\;$

$x = r\cos \theta = 8\cos \theta$, and

$y = r\sin\theta = 8\sin\theta$.

When you expand $x^2 + (y-10)^2 = 8^2$, we can use the identity $$\sin^2\theta + \cos^2 \theta = 1$$


This gives us

$$\begin{align}x^2 + (y-10)^2 = 8^2 & \iff x^2 + y^2 - 20 y + 100 = 64 \\ \\& \iff 64(\cos^2\theta + \sin^2\theta) - 20(8\sin\theta) +100 = 64 \\ \\ &\iff 20 (8\sin \theta) = 100 \\ \\ & \iff 8\sin\theta = 5 \\ \\&\iff \cdots \end{align}$$

EDIT:

No, you can't express the equation of a circle with $\sin, \cos$ in cartesian notation, and neither of your posted expressions represent the posted equation: expressing, e.g., $y$ as a function of $x$ where $x$ is in the argument of $\sin$ or $\cos$.

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  • $\begingroup$ But isn't that a parametric equation (en.wikipedia.org/wiki/Parametric_equation)? My question is how to get this in the form: $$y = ... sin(...x...)$$ $\endgroup$ – Silex Jan 21 '14 at 13:12
  • $\begingroup$ You asked whether you could express the same curve, but using $\sin$ instead. And you can. $\endgroup$ – amWhy Jan 21 '14 at 13:19
  • $\begingroup$ Sorry if you misread me but the subject explicitely says "without parametric equation". Maybe I should have said "Using cartesian equation only"? I'm not an expert in naming those things. $\endgroup$ – Silex Jan 21 '14 at 13:23
  • $\begingroup$ @Silex Good luck, and good day. $\endgroup$ – amWhy Jan 21 '14 at 13:28
  • $\begingroup$ Thanks for your answer, but note you can still answer my question. A simple "no, you can't express a circle equation with sin/cos in cartesian notation" is enough for me. I'm trying to debunk wether this y = A * sin(pi*x/2*A) has any merits $\endgroup$ – Silex Jan 21 '14 at 13:31
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Your problem is to find the area of a quarter of a circle enclosed in a circumference which is given by:

$$x^2 + (y-8)^2 = 8^2.$$

First of all, it would be useful to make a translation by defining $z = y-8$ so your circumference become:

$$x^2 + z^2 = 8^2,$$

with the same area, of course. Then, the area you request is given by:

$$A = \int \int_R 1 \, dxdy,$$

where $R$ is the region in the $(x,z)$ plane where your circle lies and it's given by:

$$R \equiv \{ (x,z) \in \mathbb{R}^2, \ 0<x<R, \ x^2 + z^2 = 8^2\}, $$

which can be rewritten more conveniently by using polar coordinates as follows:

$$ x = \rho \cos \theta, \quad y = \rho \sin \theta, \quad \rho \in [0,8], \quad \theta \in[0, \pi/2], $$

then:

$$A = \int \int_R |J(\rho,\theta)| d\rho d\theta = \int^{\pi/2}_0 \int^8_0 \rho \, d\rho d\theta = \frac{\pi 8^2}{4},$$

as you would expect.

Cheers!

I hope this is useful to you.

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  • $\begingroup$ Sorry I have just read you don't want this parametrization. $\endgroup$ – Dmoreno Jan 21 '14 at 13:36
  • $\begingroup$ No worries, thanks for the still interesting example. $\endgroup$ – Silex Jan 21 '14 at 22:59
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If you really wish to get rid of trigonometric functions, you can always expand the integrand as an infinite Taylor series built at x=0, compute the antiderivative and compute the integral. But you must understand that it is supposed to be an infinite series and that the result will then be dependent on the number of terms you use.

I took the problem as you submitted it to Wolfram Alpha and I followed the steps I described above. The antiderivative is

2 + x^2 / 16 + x^4 / 4096 + x^6 / 524288 + 5 x^8 / 268435456 + ....

Concerning the integral, depending on the number of terms, its value is successively 10.0000, 12.6042, 12.7568, 12.7780, 12.7821, 12.7830, 12.7832, 12.7833. The last number coincides with the approximate value given by Wolfram Alpha .

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  • $\begingroup$ Interesting! Thanks for the idea, it might be useful another time. $\endgroup$ – Silex Jan 21 '14 at 23:00

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