How do I solve this integral for the arc length of a cardioid?

[Telemachus]

Homework Statement

Hi there. Well, the problem statement says: Calculate the arc length of the cardioid $$\rho=a(1+\cos\theta)$$.

Homework Equations

So I used the formula for the arc length in the polar form:

$$\int_a^b \! \sqrt{(\rho)^2+(\displaystyle\frac{d\rho}{d\theta})^2} \, d\theta$$

The Attempt at a Solution

Then I get
$$\int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta$$

$$=\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta} \, d\theta$$

$$=\int_0^{2\pi} \! \sqrt{a^2(2\cos\theta+\cos^2\theta+\sin^2\theta)} \, d\theta$$

$$=\int_0^{2\pi} \! |a|\sqrt{2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta=|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta$$

I can't solve this integral, I don't know how to. Any help?

 

[Mark44]

Homework Statement

Hi there. Well, the problem statement says: Calculate the arc length of the cardioid $$\rho=a(1+\cos\theta)$$.

Homework Equations

So I used the formula for the arc length in the polar form:

$$\int_a^b \! \sqrt{(\rho)^2+(\displaystyle\frac{d\rho}{d\theta})^2} \, d\theta$$

The Attempt at a Solution

Then I get
$$\int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta$$

$$=\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\sin^2\theta} \, d\theta$$

In the equation above, you're missing a term.

$$=\int_0^{2\pi} \! \sqrt{a^2(2\cos\theta+\cos^2\theta+\sin^2\theta)} \, d\theta$$

$$=\int_0^{2\pi} \! |a|\sqrt{2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta=|a|\int_{0^2\pi} \!\sqrt{2\cos\theta+1} \, d\theta$$

I can't solve this integral, I don't know how to. Any help?

 

[Telemachus]

Sorry:
$$\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta} \, d\theta$$

Thats it.

 

[Mark44]

That's it. With some simplification, the quantity in the radical is 2a² + 2a²cos(theta). I'm not sure there's a nice, neat closed form antiderivative for this problem. Are you sure you have to actually evaluate the integral, or do you just need to set it up?

 

[Dick]

You can write 1+cos(theta) as a perfect square using a double (or half) angle formula.

 

[Telemachus]

Mm it asked for the arc length, didn't say just set it up. Mathematica gives 4 Sqrt[3] Abs[a] EllipticE[4/3] as the solution, and I don't know what the hell "EllipticE" means, less how to arrive to that result :P

The other I had of the same kind I think I've solved it right, but it was easier, it asks for the second revolution of the spiral $$\rho=e^{2\theta}$$

So I did $$\rho'=2e^{2theta}$$

$$\int_{2\pi}^{4\pi} \! \sqrt{(e^{2\theta})^2+(2e^{2\theta})^2} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{e^{4\theta}+(4e^{4\theta}} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{5e^{4\theta}} \, d\theta$$

$$\int_{2\pi}^{4\pi} \! \sqrt{5e^{4\theta}} \, d\theta$$

$$\sqrt{5}\int_{2\pi}^{4\pi} \! e^{2\theta} \, d\theta$$

$$\sqrt{5}(\dysplaystyle\frac{e^{8\pi}-e^{4\pi}}{2})$$

Is that right?

 

[Mark44]

Mathematica is saying that the first integral is an elliptic integral (see here).

Your work for the second integral looks fine.

 

[Telemachus]

You can write 1+cos(theta) as a perfect square using a double (or half) angle formula.

I think I can use the half angle formula (I watched at wikipedia after your post).
This:
$$\cos \frac{\theta}{2} = \pm\, \sqrt{\frac{1 + \cos\theta}{2}}$$

Am I right? thanks Dick

Mathematica is saying that the first integral is an elliptic integral (see here).

Your work for the second integral looks fine.

Thanks Mark

 

[Dick]

I think I can use the half angle formula (I watched at wikipedia after your post).
This:
$$\cos \frac{\theta}{2} = \pm\, \sqrt{\frac{1 + \cos\theta}{2}}$$

Am I right? thanks Dick

That's the one.

 

[Telemachus]

Thanks both of you. See you later!

 

[Telemachus]

mm It seems it won't help, I've tried factorizing the 2, but...

$$|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta=\sqrt{2}|a|\int_0^{2\pi} \!\sqrt{\cos\theta+\dyslpaystyle\frac{1}{2}}$$

I thought I was on the right way, but I've missed that the denominator 2 its only for the 1 inside the square root.

And the double angle formula I think it's not going to work neither

$$\begin{align}<br /> \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= \color{red} 2 \cos^2 \theta - 1 \color{black} \\ <br /> &= 1 - 2 \sin^2 \theta \\ &= \frac{1 - \tan^2 \theta} {1 + \tan^2 \theta}<br /> \end{align}$$

It seems to me that Mark were right :P

 

[vela]

You're using the integrand with the mistake in it. (You can tell it's wrong because the radicand is negative for some values of theta.) Use the correct one you found later, and you'll be able to integrate it using Dick's hint.

 

[Telemachus]

I don't think so, I can't see the mistake. I've already corrected the one that Mark marked :P

 

[Mark44]

The integral should look like this:
$$=\sqrt{2}|a|\int_0^{2\pi} \sqrt{1 + cos\theta} d\theta$$

Use Dick's hint.

 

[vela]

$$=\int_0^{2\pi} \! |a|\sqrt{2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta=|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta$$

That was your original integral.

Sorry:
$$\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta} \, d\theta$$

And that was your corrected one.

$$|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta=\sqrt{2}|a|\int_0^{2\pi} \!\sqrt{\cos\theta+\dyslpaystyle\frac{1}{2}}$$

Now you are using the original one again, which has the mistake.

 

[Mark44]

vela,
This is the original integral:
$$\int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta$$

The radicand simplifies to
$$\sqrt{2a^2(1 + cos(\theta))}$$

 

[Telemachus]

Right, I had a mistake there. Thanks both. So the integral keeps this way:

$$\sqrt{2}|a|\int_0^{2\pi} \! \sqrt{1+\cos\theta} \, d\theta$$ Dividing and multiplying by two I can get something to apply the half angle formula as Dick said before.