Arclength of polar cardiod problem

[QuarkCharmer]

Homework Statement

My textbook sets up the integral, but does not solve, claiming that it's "trivial to solve manually or by using a CAS". I put the integral into my TI-89, and sure enough, there is a solution, and that solution happens to be "8". However...

Homework Equations

The actual problem is to find the total arclength of r = 1+sin(x)
(It's polar, I just let theta = x, since theta is a pain to LaTeX)

The Attempt at a Solution

Here is my attempt at solving this integral manually:

s = \int_{0}^{2π}\sqrt{(1+sin^{2}(x))+cos(x)^{2}}dx

\int_{0}^{2π}\sqrt{2+2sin(x)}dx

\int_{0}^{2π}\sqrt{2+2sin(x)} \frac{\sqrt{2-2sin(x)}}{\sqrt{2-2sin(x)}}dx

\int_{0}^{2π}\sqrt{\frac{4-4sin^{2}(x)}{2-2sin(x)}}dx

\int_{0}^{2π}\sqrt{\frac{4cos^{2}(x)}{2(1-sin(x))}}dx

\sqrt{2}\int_{0}^{2π}\frac{cos(x)}{\sqrt{1-sin(x)}}dx

u = 1-sin(x)
du = -cos(x)dx

-\sqrt{2}\int_{1}^{1}\frac{1}{\sqrt{u}}du

I stopped here, because it's clear to me that the integral from one to one will be zero...

...and now the whole thing equals zero... I have tried doing the integral with substitutions other than u=1-sin(x) to no avail. I tried to devise a way so I could take a limit (much like some improper integral problems are done), no luck. I can't seem to figure this out without using a CAS! It is one thing if I just got to a point where I didn't know how to proceed I could deal with, but I come up with zero every time!

This is my best attempt, I have done it 11 different ways now..

 

[Dick]

You did a pretty dangerous thing there. You said sqrt(cos(x)^2)=cos(x). That's not right. cos(x) is negative for some values in [0,2pi]. It's |cos(x)| isn't it?

 

[QuarkCharmer]

Yeah, that makes sense, good catch. I still have no clue how to proceed though, especially now with the abs in the integral

\sqrt{2}\int_{0}^{2π}\frac{|cos(x)|}{\sqrt{1-sin(x)}}dx

 

[Dick]

Yeah, that makes sense, good catch. I still have no clue how to proceed though, especially now with the abs in the integral

\sqrt{2}\int_{0}^{2π}\frac{|cos(x)|}{\sqrt{1-sin(x)}}dx

Split the region of integration up into parts where cos(x) has a definite sign. Isn't that what you usually do with abs problems? It's premature to cry over it.

 

[QuarkCharmer]

\sqrt{2}\int_{\frac{-π}{2}}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx

Since, cos(x) will be positive in that domain.

And;

\sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx

Like that? Forgive me, I have never encountered an integral with an abs yet.

So:

\sqrt{2}\int_{\frac{-π}{2}}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx

 

[Dick]

\sqrt{2}\int_{\frac{-π}{2}}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx

Since, cos(x) will be positive in that domain.

And;

\sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx

Like that? Forgive me, I have never encountered an integral with an abs yet.

So:

\sqrt{2}\int_{\frac{-π}{2}}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx

No, you want to integrate from 0 to 2pi. Integrate separately over the regions [0,pi/2], [pi/2,3pi/2] and [3pi/2,2pi] and add them up with appropriate signs.

 

[QuarkCharmer]

Thanks for the help.

No, you want to integrate from 0 to 2pi. Integrate separately over the regions [0,pi/2], [pi/2,3pi/2] and [3pi/2,2pi] and add them up with appropriate signs.

<br /> \sqrt{2}\int_{0}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{3π}{2}}^{2π}\frac{cos(x)}{\sqrt{1-sin(x)}}dx<br />

(No clue why that sqrt is not showing.)

I see that you are splitting it up for the regions in which cos(x) would be positive or negative, but I am not sure how to sign the cos(x) in the functions. If integrating through the range of pi/2 to 3pi/2, cos(x) would turn out negative anyway? I am not seeing what effect this would have exactly?

Here is how I am imagining the situation:

Suppose you had \int_{0}^{2π}|sin(x)|dx.

Since, that integral would be zero if there were no abs value, to handle the abs value case I would split the integral up like so:

\int_{0}^{π}sin(x)dx + \int_{π}^{2π}-sin(x)dx

In this way, when sin(x) is negative, the negative that I placed into the split region would force the sin(x) to act positive?

 

[Dick]

Thanks for the help.



<br /> \sqrt{2}\int_{0}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{3π}{2}}^{2π}\frac{cos(x)}{\sqrt{1-sin(x)}}dx<br />

(No clue why that sqrt is not showing.)

I see that you are splitting it up for the regions in which cos(x) would be positive or negative, but I am not sure how to sign the cos(x) in the functions. If integrating through the range of pi/2 to 3pi/2, cos(x) would turn out negative anyway? I am not seeing what effect this would have exactly?

Here is how I am imagining the situation:

Suppose you had \int_{0}^{2π}|sin(x)|dx.

Since, that integral would be zero if there were no abs value, to handle the abs value case I would split the integral up like so:

\int_{0}^{π}sin(x)dx + \int_{π}^{2π}-sin(x)dx

In this way, when sin(x) is negative, the negative that I placed into the split region would force the sin(x) to act positive?

That's it exactly. Now evaluate those integrals and add them. I get 8.