# Is it possible to find the arclength of semicircle sqrt(16-x^2) w/o trig sub?
**URL:** https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414
**Category:** Factual Questions
**Created:** [March 3, 2011, 10:32pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414 "2011-03-03T22:32:06Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@miragesyzygy](https://boards.straightdope.com/u/miragesyzygy)
#### Post date: [March 3, 2011, 10:32pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/1 "2011-03-03T22:32:06Z")
not a homework problem,
I know it can be done with a trignometric substituion involving sin, but if a class hasn’t covered that yet, how would you begin to integrate the mess resolving from that?
To find the Arc Length, I did
Integral from -4 to 4 of
sqrt(1+ ((d/dx)(sqrt(16-x^2)))^2)
(16-x^2) ^1/2
1/2 (16-x^2)^-1/2 X -2x
then square that, add 1, and then integrate?
it looks even messier in this textbox, but did I err somewhere, or does using trig sub (which we havn’t learned yet) make life a lot easier?
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### Author:  [@Sunspace](https://boards.straightdope.com/u/Sunspace)
#### Post date: [March 3, 2011, 10:54pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/2 "2011-03-03T22:54:17Z")
I need more information. I’m relearning this stuff, and I feel as if I’m kind of plunked down in the middle of things. What is x defined as again?
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### Author:  [@Dervorin](https://boards.straightdope.com/u/Dervorin)
#### Post date: [March 3, 2011, 11:16pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/3 "2011-03-03T23:16:25Z")
Um. Perhaps I’m being rather dense here, but isn’t the arc length of a semi-circle pi \* r by definition, where r is the radius?
r = 4, arc length = 4 \* pi.
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [March 3, 2011, 11:16pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/4 "2011-03-03T23:16:27Z")
Can’t you just use good old 2pi r (or pi r, since it’s a semicircle) here?
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### Author:  [@Sunspace](https://boards.straightdope.com/u/Sunspace)
#### Post date: [March 3, 2011, 11:47pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/5 "2011-03-03T23:47:45Z")
Yeah, I’m wondering whether the OP is trying to integrate off an element of area between two radius bounds across a subtended angle, but I can’t really tell. When I wanted to find the area of a sector of a circular band, I just used area of a sector = 1/2 \* (subtended angle in radians) \* radius[sup]2[/sup], and subtracted the smaller sector from the larger.
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### Author:  [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)
#### Post date: [March 4, 2011, 12:55am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/6 "2011-03-04T00:55:30Z")
Are you looking for a way to evaluate the integral as given, without using trig substitution? If so, you could use numerical integration (e.g. Simpson’s Rule) or look up the antiderivative on a sufficiently detailed table of integrals, if these approaches aren’t “out of bounds” for what you’re doing.
Or are you looking for a way to derive the arclength of the given curve? Does the class in question know about parametric equations, or polar coordinates? Using either of those, it turns into an easy-to-evaluate integral.
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### Author:  [@glowacks](https://boards.straightdope.com/u/glowacks)
#### Post date: [March 4, 2011, 1:22am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/7 "2011-03-04T01:22:08Z")
If it’s solvable using a trig substitution, that’s probably the only way to do it analytically with only one coordinate change. But it probably would be greatly simplified by transforming to polar coordinates where the arc of a circle has a much simpler representation, and certainly if all you need is a number it can be solved numerically nearly instantaneously once you figure out how to tell the computer software package to find out what you want to know.
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### Author:  [@MikeS](https://boards.straightdope.com/u/MikeS)
#### Post date: [March 4, 2011, 1:47am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/8 "2011-03-04T01:47:20Z")
Yeah, I’m not sure that you’re going to get it without appealing to trigonometric functions. We know that the answer is 4π, and I really don’t see how you’re going to get a multiple of π out of polynomials and square roots without detouring through trig.
Note: this is not a proof, just a gut feeling.
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### Author:  [@miragesyzygy](https://boards.straightdope.com/u/miragesyzygy)
#### Post date: [March 4, 2011, 3:06am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/9 "2011-03-04T03:06:28Z")
evaluate the integral as given
we only have the basic integration rules, usub, and trignometric inverses (but not substitutions). if sqrt(16-x^2) were in the denominator, I can use arcsin, but not when it’s by itself.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 4, 2011, 4:07am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/10 "2011-03-04T04:07:24Z")
> [@MikeS](#):
>
> Yeah, I’m not sure that you’re going to get it without appealing to trigonometric functions. We know that the answer is 4π, and I really don’t see how you’re going to get a multiple of π out of polynomials and square roots without detouring through trig.
That’s my intuition as well.
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### Author:  [@Xema](https://boards.straightdope.com/u/Xema)
#### Post date: [March 4, 2011, 5:53am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/11 "2011-03-04T05:53:12Z")
> [@MikeS](#):
>
> I really don’t see how you’re going to get a multiple of π out of polynomials and square roots without detouring through trig.
I’m not sure how useful it might be, but you can get pi from this series:
pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 …
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [March 4, 2011, 7:17am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/12 "2011-03-04T07:17:11Z")
First of all, as others have pointed out, calculating the arclength you are looking at is clearly calculating the arclength of a semicircle of radius 4, which is _definitionally_ 4π, no work required.
That having been said:
> [@miragesyzygy](#):
>
> evaluate the integral as given
>
> we only have the basic integration rules, usub, and trignometric inverses
Well, then, you have “trig substitution” as well, whether you were told so or not. “Trig substitution” is just the special case of “u-substitution” where you happen to take u to be an inverse trigonometric function of x…
> [@](#):
>
> if sqrt(16-x^2) were in the denominator, I can use arcsin, but not when it’s by itself.
Take another look at your integrand: you _do_ have sqrt(16 - x^2) in the denominator…
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [March 4, 2011, 7:32am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/13 "2011-03-04T07:32:24Z")
> [@Thudlow\_Boink](#):
>
> Does the class in question know about parametric equations, or polar coordinates? Using either of those, it turns into an easy-to-evaluate integral.
> [@glowacks](#):
>
> If it’s solvable using a trig substitution, that’s probably the only way to do it analytically with only one coordinate change. But it probably would be greatly simplified by transforming to polar coordinates where the arc of a circle has a much simpler representation
Transforming into polar _is_ trig substitution, in all but name: taking x = 4 cos(theta) and going from there.
[Or, more generally, performing the multivariable substitution \ = r \ and going from there. If substitution where the old variables are trigonometric functions of the new ones isn’t trig substitution, then what is?]
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### Author:  [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)
#### Post date: [March 4, 2011, 12:59pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/14 "2011-03-04T12:59:35Z")
The OP is incoherent. Sqrt(16 - x^2) is **not** a circle; it is an expression. For example, when x = 1, it evaluates to sqrt(15) and so on.
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### Author:  [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)
#### Post date: [March 4, 2011, 3:19pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/15 "2011-03-04T15:19:02Z")
> [@Indistinguishable](#):
>
> Take another look at your integrand: you _do_ have sqrt(16 - x^2) in the denominator…
Yeah, actually, you’re right.
This is worked out as [Example 1 here](http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/arclength.html).
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### Author:  [@miragesyzygy](https://boards.straightdope.com/u/miragesyzygy)
#### Post date: [March 4, 2011, 10:59pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/16 "2011-03-04T22:59:44Z")
No I think the sqrt in the integrand becomes squared, thus eliminating the sqrt, since the formula for arc length involves a square of the derivative, and not the derivative itself.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 4, 2011, 11:47pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/17 "2011-03-04T23:47:05Z")
> [@miragesyzygy](#):
>
> No I think the sqrt in the integrand becomes squared, thus eliminating the sqrt, since the formula for arc length involves a square of the derivative, and not the derivative itself.
Work it out yourself and check this.
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### Author:  [@Manlob](https://boards.straightdope.com/u/Manlob)
#### Post date: [March 5, 2011, 5:31am UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/18 "2011-03-05T05:31:22Z")
Integral of 4/sqrt(16-x^2) from -4 to 4…
Make the change of variable u=(i_x+sqrt(16-x^2)/4 where i=sqrt(-1)
For brevity, I will use S to represent the function of x: S=sqrt(16-x^2), so u=i_x+S and du=-u/(i_S). The integrand, 4/S, expressed in terms of u is: -4i_du/du, which is easily integrated to give -4i_ln(u) = -4i_ln((i\*x+sqrt(16-x^2))/4)
Plugging in the limits of integration (-4 to 4) gives: -4_i_ln(i)+4_i_ln(-i). Since ln(i)=i_pi/2 and ln(-i)=-i_pi/2, the result is 4\*pi.
Another approach would be to expand the integrand in an infinite series and integrate term by term to get a result with a recognizable series for pi. A well known series for pi is the one Xema mentioned above. To get that one, before expanding the integrand in a series, make the change of variable t=sqrt(4-x)/sqrt(4+x). This gives an integral that could be solved with arctan(t) or expanded in a series and integrated term by term.
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### Author:  [@miragesyzygy](https://boards.straightdope.com/u/miragesyzygy)
#### Post date: [March 5, 2011, 5:50pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/19 "2011-03-05T17:50:36Z")
but it’s not 4 over that, it’s just the arc length of
y = sqrt(16-x^2) from -4 to 4
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [March 5, 2011, 7:39pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/20 "2011-03-05T19:39:28Z")
> [@miragesyzygy](#):
>
> No I think the sqrt in the integrand becomes squared, thus eliminating the sqrt, since the formula for arc length involves a square of the derivative, and not the derivative itself.
It also involves a square root again, which brings you back to where you want to be. Look:
> [@miragesyzygy](#):
>
> To find the Arc Length, I did
>
> Integral from -4 to 4 of
>
> **sqrt(1+ ((d/dx)(sqrt(16-x^2)))^2)**
>
> (16-x^2) ^1/2
>
> 1/2 (16-x^2)^-1/2 X -2x
>
> then square that, add 1, and then integrate?
Inbetween “square that, add 1” and “then integrate”, you need to take a square root, as noted in the bolded portion of the enlarged line.
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