# 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:** 1\
**Page:** 2

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**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\
**Post date:** [March 5, 2011, 7:57pm UTC](https://boards.straightdope.com/t/is-it-possible-to-find-the-arclength-of-semicircle-sqrt-16-x-2-w-o-trig-sub/573414/21 "2011-03-05T19:57:05Z")

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> [@miragesyzygy](#):
>
> 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?

Have you been given a table, or access to a table, of integrals?

> [@Hari\_Seldon](#):
>
> 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.

But, f(x) = Sqrt(16 - x^2), or y = Sqrt(16 - x^2), are circles. That appears to be what the OP is about.

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