# How do I solve this integral?

Printable View

• Aug 4th 2010, 03:22 PM
Ulysses
How do I solve this integral?
Hi there. Well, the problem statement says: Calculate the arc length of the cardioid $\rho=a(1+\cos\theta)$.

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

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

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?
• Aug 4th 2010, 04:48 PM
skeeter
Quote:

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

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

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

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

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

$=\int_0^{2\pi} \! \sqrt{a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta) } \, d\theta$ ... algebra error here (fixed)

...
• Aug 4th 2010, 05:57 PM
Ulysses
Thanks skeeter.