Is it possible to find the arclength of semicircle sqrt(16-x^2) w/o trig sub?

General Questions
#1

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?

#2

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?

#3

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.

#4

Can’t you just use good old 2pi r (or pi r, since it’s a semicircle) here?

#5

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.

#6

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.

#7

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.

#8

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.

#9

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.

#10

That’s my intuition as well.

#11

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 …

#12

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:

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…

Take another look at your integrand: you do have sqrt(16 - x^2) in the denominator…

#13

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 <x, y> = r <cos(theta), sin(theta)> and going from there. If substitution where the old variables are trigonometric functions of the new ones isn’t trig substitution, then what is?]

#14

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.

#15

Yeah, actually, you’re right.

This is worked out as Example 1 here.

#16

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.

#17

Work it out yourself and check this.

#18

Integral of 4/sqrt(16-x^2) from -4 to 4…

Make the change of variable u=(ix+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=ix+S and du=-u/(iS). The integrand, 4/S, expressed in terms of u is: -4idu/du, which is easily integrated to give -4iln(u) = -4iln((i*x+sqrt(16-x^2))/4)

Plugging in the limits of integration (-4 to 4) gives: -4iln(i)+4iln(-i). Since ln(i)=ipi/2 and ln(-i)=-ipi/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.

#19

but it’s not 4 over that, it’s just the arc length of

y = sqrt(16-x^2) from -4 to 4

#20

It also involves a square root again, which brings you back to where you want to be. Look:

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.