# Polar Arc length

1. Homework Statement

Find the length pf the curve over the given interval.
$$r=1+\sin\theta$$
$$0\preceq\theta\preceq\2\pi$$

3. The Attempt at a Solution
Ok I set it up as:
$$2\pi$$
$$\int\sqrt((1+\sin\theta)^2+cos^2\theta)$$
0

and by simplifying and integrating, I get

$$2\pi$$
$$-2\sqrt2[\sqrt(1-\sin\theta)]$$
0

$$-2\sqrt2[(1-0)-(1-0)] =0$$

and obviously it is wrong,

I check the solution it has the same everything but the range ,
it obviously broke down the whole length into 2 times 1 piece from $$\pi/2 to 3\pi/2$$ and the answer is 8

My question is why I get zero within my range, and why broke it down into the range above?

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Dick
Homework Helper
You might wish to elaborate about how you got that integral. It's not what I get. You want to do the integral using a double angle formula to write the integrand as a perfect square. It also may help to shift the limits of integration by using sin(x+pi/2)=cos(x). You should also remember that sqrt(x^2)=abs(x) for any expression x. So to integrate something like that you'll need to break into regions where 'x' is positive.

You might wish to elaborate about how you got that integral. It's not what I get. You want to do the integral using a double angle formula to write the integrand as a perfect square. It also may help to shift the limits of integration by using sin(x+pi/2)=cos(x). You should also remember that sqrt(x^2)=abs(x) for any expression x. So to integrate something like that you'll need to break into regions where 'x' is positive.
There is some mistake on my first post but anyway I integrate it like this: (ignore the limits for the moment)

$$\int\sqrt((1+\sin\theta)^2+cos^2\theta)$$

= $$\sqrt2 \int\sqrt((1+\sin\theta)$$

= $$\sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)}$$
(multiply up and down by $$\sqrt(1-\sin\theta)$$

= $$2\sqrt2 [ \sqrt((1-\sin\theta)]$$ (by substitution)

which if I use the limit $$\pi/2$$ to $$3\pi/2$$ and times 2
I get the answer, by I still don't understand what decision it based on setting the limits:uhh:,
I also tried from 0 to pi and times 2 and it came out to be zero :yuck:, also when I trace on the calculator ( $$\pi/2$$ to $$3\pi/2$$)the x's are all -ve :uhh: clue?

So yes why the selected limits Also it will be great if you can demonstrate me a different way of integrating it and what do you mean by to shift the limits of integration by using sin(x+pi/2)=cos(x) Thanks a million~

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Ok wait I am up to something, it is because of the behaviour of the sin curve ?

for not getting zero I have to use that certain limit? :rofl:

http://hk.geocities.com/ymtsang2606/sin.jpg [Broken]

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Dick
Homework Helper
There is some mistake on my first post but anyway I integrate it like this: (ignore the limits for the moment)

$$\int\sqrt((1+\sin\theta)^2+cos^2\theta)$$

= $$\sqrt2 \int\sqrt((1+\sin\theta)$$

= $$\sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)}$$
(multiply up and down by $$\sqrt(1-\sin\theta)$$

= $$2\sqrt2 \int \sqrt((1-\sin\theta)$$ (by substitution)

which if I use the limit $$\pi/2$$ to $$3\pi/2$$ and times 2
I get the answer, by I still don't understand what decision it based on setting the limits:uhh:,
I also tried from 0 to pi and times 2 and it came out to be zero :yuck:, also when I trace on the calculator ( $$\pi/2$$ to $$3\pi/2$$)the x's are all -ve :uhh: clue?

So yes why the selected limits Also it will be great if you can demonstrate me a different way of integrating it and what do you mean by to shift the limits of integration by using sin(x+pi/2)=cos(x) Thanks a million~
That's a clever trick. But it's hiding something from you. sqrt(cos(theta)^2)=abs(cos(theta)). As theta goes from 0 to pi, cos changes sign. Integrate from 0 to pi/2, then from pi/2 to pi and add the absolute values.

That's a clever trick. But it's hiding something from you. sqrt(cos(theta)^2)=abs(cos(theta)). As theta goes from 0 to pi, cos changes sign. Integrate from 0 to pi/2, then from pi/2 to pi and add the absolute values.
How do u get to sqrt(cos(theta)^2)?

Dick
Homework Helper
sqrt(1-sin(theta))*sqrt(1+sin(theta))=sqrt((1-sin(theta))*(1+sin(theta))=
sqrt(1-sin(theta)^2)=sqrt(cos(theta)^2). Isn't that what you did?

$$\int\sqrt((1+\sin\theta)^2+cos^2\theta)$$

= $$\sqrt2 \int\sqrt((1+\sin\theta)$$ (then I multiply up and down by $$\sqrt(1-\sin\theta)$$ ) <-- do u mean the sqrt cos^2theta on the top after multiplying?

= $$\sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)}$$

= $$2\sqrt2[\sqrt((1-\sin\theta)]$$

Dick
Homework Helper
Yes, I mean the cos(theta)^2 on the top under the square root.

Yes, I mean the cos(theta)^2 on the top under the square root.

Ok I get it now, so when you integrate you must make sure that the signs of the function would not change through out the limit range , cause otherwise they will offset each other, and given this case :$$\sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)}$$

the sqrtcos^2theta on the numerator change signs with the range 0 to pi so you have to break it down in to two parts and integrate, while sin doen't change signs in the denominator within 0 to pi so it doen't matter.

There is some mistake on my first post but anyway I integrate it like this: (ignore the limits for the moment)

$$\int\sqrt((1+\sin\theta)^2+cos^2\theta)$$

= $$\sqrt2 \int\sqrt((1+\sin\theta)$$

= $$\sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)}$$
(multiply up and down by $$\sqrt(1-\sin\theta)$$

= $$2\sqrt2 \int \sqrt((1-\sin\theta)$$ (by substitution)

which if I use the limit $$\pi/2$$ to $$3\pi/2$$ and times 2
I get the answer, by I still don't understand what decision it based on setting the limits:uhh:,
I also tried from 0 to pi and times 2 and it came out to be zero :yuck:, also when I trace on the calculator ( $$\pi/2$$ to $$3\pi/2$$)the x's are all -ve :uhh: clue?

So yes why the selected limits Also it will be great if you can demonstrate me a different way of integrating it and what do you mean by to shift the limits of integration by using sin(x+pi/2)=cos(x) Thanks a million~

There's a superfluous integral sign in that post (the last line, where you've already performed the integration)... sorry to nitpick, but it might be confusing for those who are first signing on.