# Find circumference of cardioid

• October 11th 2010, 05:30 PM
arze
Find circumference of cardioid
The cardioid $r=a(1-\sin\theta)$; find the length of its circumference.

The formula for polar equations:
$\displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta$

$\frac{dr}{d\theta}=-a\cos\theta$
$\displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}} d\theta$
$\displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra c{1}{2}}$
$\displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta$

$(1-\sin\theta)^{1/2}$ is what stumps me, how do i convert that to something that can be integrated?
Thanks
• October 11th 2010, 07:45 PM
mr fantastic
Quote:

Originally Posted by arze
The cardioid $r=a(1-\sin\theta)$; find the length of its circumference.

The formula for polar equations:
$\displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta$

$\frac{dr}{d\theta}=-a\cos\theta$
$\displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}} d\theta$
$\displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra c{1}{2}}$
$\displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta$

$(1-\sin\theta)^{1/2}$ is what stumps me, how do i convert that to something that can be integrated?
Thanks

integrate Sqrt&#91;1 - Sin&#91;x&#93;&#93; - Wolfram|Alpha

Click on Show steps.
• October 11th 2010, 07:56 PM
arze
Quote:

Originally Posted by mr fantastic

Ok, so i did that, and got:
$2a\sqrt{2} [2\sqrt{\sin\theta+1}]^{\pi}_0$
which gives me 0.
I tried adding the integrals between $\frac{\pi}{2}$ and 0, and $\pi$ and $\frac{\pi}{2}$ they still add to 0.
Thanks!
• October 11th 2010, 08:18 PM
Soroban
Hello, arze!

Quote:

The cardioid $r=a(1-\sin\theta)$; find the length of its circumference.

$\displaystyle s \;=\;2\sqrt{2}\,a\int^{\pi}_0\left(1-\sin\theta\right)^{\frac{1}{2}}d\theta$

We have: . $\sqrt{1-\sin\theta}$

Multiply by $\frac{\sqrt{1+\sin\theta}}{\sqrt{1+\sin\theta}}:\; \;\dfrac{\sqrt{1-\sin\theta}}{1}\cdot\dfrac{\sqrt{1+\sin\theta}}{\s qrt{1+\sin\theta}} \;=\;\dfrac{\sqrt{1-\sin^2\theta}}{\sqrt{1+\sin\theta}}$

. . . . . . . . . . . . . $=\;\dfrac{\sqrt{\cos^2\theta}}{\sqrt{1+\sin\theta} } \;=\;\dfrac{\cos\theta}{\sqrt{1+\sin\theta}}$

$\text{The integral becomes: }\;\displaystyle 2\sqrt{2}\,a\int (1 + \sin\theta)^{-\frac{1}{2}}\,\cos\theta\,d\theta$
$\text}Let }\,u \:=\:1 + \sin\theta$

Got it?
• October 11th 2010, 08:19 PM
arze
yes, got it!
Thanks!
• October 11th 2010, 08:34 PM
arze
ok, now i have another problem. between pi and 0, u has the same values, what values should i use? for the angles?