proof of area of circle

aron101782

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I have a question about the derivation of the formula for the area of a circle to be pi r^2 .
When substituting for x I have used both r sint and r cost and both seem to work. However when using what would at first seem to be obviously correct x=rcost leads to a negative integrand and 0 as an upper bound of integration. My question is that is this the only reason why rsint is chosen as an arbitrary substitution.

-Aron
 
I have a question about the derivation of the formula for the area of a circle to be pi r^2 .
When substituting for x I have used both r sint and r cost and both seem to work. However when using what would at first seem to be obviously correct x=rcost leads to a negative integrand and 0 as an upper bound of integration. My question is that is this the only reason why rsint is chosen as an arbitrary substitution.

-Aron

Why not show us your integral?
 
So we have a circle of radius r centered at the origin

x^2+y^2=r^2
y=+-(r^2-x^2)^1/2

Here we will need only focus on the area under the curve in the first quadrant, such that:

4I(0, r) (r^2-x^2)^1/2 dx

x=rsint

at the end of the integrals we solve for dx, such that:

dx/dt=rcost
dx=rcost dt

we then replace the parameters of integration, such that:

when x=0 then t=0
when x=r then t=pi/2

therefore:

4I(0, pi/2) r(1-sin^2t)^1/2 rcost dt
4r^2 I(0, pi/2) cos^2t dt

We will need to substitute cos^2t with (1+cos2t)/2, such that:
(See this document for the proof https://www.themathpage.com/aTrig/sum-proof.htm )

2r^2I(0, pi/2)1+cos2t dt
2r^2(t+sin2t/2)[0, pi/2]
pi r^2 here is our formula for the area of a circle

My question is why is this the standard trigonometric substitution. In this case instead of using x=rsint it seems much more reasonable to use x=rcost. It does lead to the same solution. Although, the integral is a bit more messy.

-Aron
 
So we have a circle of radius r centered at the origin

x^2+y^2=r^2
y=+-(r^2-x^2)^1/2

Here we will need only focus on the area under the curve in the first quadrant, such that:

4I(0, r) (r^2-x^2)^1/2 dx

x=rsint

at the end of the integrals we solve for dx, such that:

dx/dt=rcost
dx=rcost dt

we then replace the parameters of integration, such that:

when x=0 then t=0
when x=r then t=pi/2

therefore:

4I(0, pi/2) r(1-sin^2t)^1/2 rcost dt
4r^2 I(0, pi/2) cos^2t dt

We will need to substitute cos^2t with (1+cos2t)/2, such that:
(See this document for the proof https://www.themathpage.com/aTrig/sum-proof.htm )

2r^2I(0, pi/2)1+cos2t dt
2r^2(t+sin2t/2)[0, pi/2]
pi r^2 here is our formula for the area of a circle

My question is why is this the standard trigonometric substitution. In this case instead of using x=rsint it seems much more reasonable to use x=rcost. It does lead to the same solution. Although, the integral is a bit more messy.

-Aron
You are correct that it does not matter. Whenever I teach this material I always have to look at the substitution that my textbook uses as I do not memorize these formulas but rather I look at a right triangle (in this case) that has one side being (r^2-x^2)^1/2 and all is good from there. As you said you can put (r^2-x^2)^1/2 on either leg. So the other leg get x and the hyp gets r. If you drew the triangle I do not think you would have ever written (1-sin^2t)^1/2. Try drawing the triangle and see how that goes. If you have sqrt(a2-x2), the sqrt of the positive term (which is a2) goes on the hyp. The sqrt of the negative term (which is x2) goes on one leg and sqrt(a2-x2) goes on the other leg. Same procedure if you had x2-a2. If you had x2+a2. one leg gets a, the other leg get x and the hypotenuse gets sqrt(x2+a2) (since both terms are positive).
 
My question is why is this the standard trigonometric substitution. In this case instead of using x=rsint it seems much more reasonable to use x=rcost. It does lead to the same solution. Although, the integral is a bit more messy.

It seems to me that you answered your own question. Why is it "much more reasonable" to use the method that leads to a little more "messy" work, when both are otherwise equivalent? The main difference is in signs, and the more negatives in a problem, the greater the risk of error. Why take that risk, when you can avoid it?

If you're right that the sine substitution is "standard" (and I don't know that any standards organization has decreed that), then there seems to be good reason for it.

But why does it matter? If you like something that isn't "standard" (that is, more popular?), but which is perfectly valid, then do what you prefer.
 
It seems to me that you answered your own question. Why is it "much more reasonable" to use the method that leads to a little more "messy" work, when both are otherwise equivalent? The main difference is in signs, and the more negatives in a problem, the greater the risk of error. Why take that risk, when you can avoid it?

If you're right that the sine substitution is "standard" (and I don't know that any standards organization has decreed that), then there seems to be good reason for it.

But why does it matter? If you like something that isn't "standard" (that is, more popular?), but which is perfectly valid, then do what you prefer.

x=asint does appear to be the standard trigonometric substitution for the expression (a^2-x^2)^1/2
 
x=asint does appear to be the standard trigonometric substitution for the expression (a^2-x^2)^1/2

I'm not really doubting that, just pointing out that there is no official designation of what is "standard", and no one to ask about the reasons.

You've just found that it is the most common method taught, and that there is some reason to prefer it, even though it doesn't matter much. Sometimes what is commonly taught is entirely arbitrary, just because teachers started copying one another after one of them happened to choose a particular option. (An example of that is writing "y = mx + b" as everyone does in America, though in other countries different letters are used.)

I'd still like to know why you think the alternative is "much more reasonable". Maybe you have an explanation I'd agree with.
 
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