Polar Arc length

1. Apr 6, 2008

johndoe

1. The problem statement, all variables and given/known data

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?

Last edited: Apr 6, 2008
2. Apr 6, 2008

Dick

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.

3. Apr 7, 2008

johndoe

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~

Last edited: Apr 7, 2008
4. Apr 7, 2008

johndoe

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:

5. Apr 7, 2008

Dick

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.

6. Apr 7, 2008

johndoe

How do u get to sqrt(cos(theta)^2)?

7. Apr 7, 2008

Dick

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?

8. Apr 7, 2008

johndoe

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

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

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

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

9. Apr 7, 2008

Dick

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

10. Apr 7, 2008

johndoe

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.

11. Apr 7, 2008

BrendanH

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.