Length of Polar Arc: 0 to 2π

[johndoe]

Homework Statement

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

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?

 

[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.

 

[johndoe]

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~

 

[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:

http://hk.geocities.com/ymtsang2606/sin.jpg

 

[Dick]

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.

 

[johndoe]

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]

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?

 

[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?

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


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

 

[Dick]

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

 

[johndoe]

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 into two parts and integrate, while sin doen't change signs in the denominator within 0 to pi so it doen't matter.

 

[BrendanH]

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.