Teacher needs help with polar co-ordinates

• Nov 4th 2007, 03:05 AM
TheBrain
Teacher needs help with polar co-ordinates
Haha. I bet this will be one of the more unusual questions you get on here. I am a trainee teacher and I noticed last week that one of the other teachers was asked a question about polar co-ordinates. He couldn't do it, and after showing it to the rest of us in the staffroom, we decided we couldn't either! Any help or guidance you could give would be very appreciated. By next week the student will probably be on the next chapter, but for the sake of our sanity we would like to see how to do it!

The question is:

Integrate (a^2) / (1 + cos theta) between 0 and pi/2 where a is just a constant.

Now we know that an integration in polar co-ordinates is

A = (a^2) / 2 x integral 1 / (1+cos theta)^2 d-theta

Everything we try from there onwards just makes things much worse. Substitution doesn't help, so I was thinking there could be some trig-identity which would make things easier but I couldn't see one.

• Nov 4th 2007, 03:37 AM
kalagota
Quote:

Originally Posted by TheBrain
Haha. I bet this will be one of the more unusual questions you get on here. I am a trainee teacher and I noticed last week that one of the other teachers was asked a question about polar co-ordinates. He couldn't do it, and after showing it to the rest of us in the staffroom, we decided we couldn't either! Any help or guidance you could give would be very appreciated. By next week the student will probably be on the next chapter, but for the sake of our sanity we would like to see how to do it!

The question is:

Integrate (a^2) / (1 + cos theta) between 0 and pi/2 where a is just a constant.

Now we know that an integration in polar co-ordinates is

A = (a^2) / 2 x integral 1 / (1+cos theta)^2 d-theta

Everything we try from there onwards just makes things much worse. Substitution doesn't help, so I was thinking there could be some trig-identity which would make things easier but I couldn't see one.

sir,
$\displaystyle \int_0^{\frac{\pi}{2}} \frac {a^2}{1+cos\theta}d\theta$

$\displaystyle a^2$ is a constant hence it can go out in the integral sign.. if you multiply it by $\displaystyle \frac{1-cos\theta}{1-cos\theta}$ i believe you'll get a result..
• Nov 4th 2007, 03:40 AM
TheBrain
Sorry kaalgota maybe I wasn't clear enough or I am still confused. I thought that when you integrate something in polar co-ordinates you have to square the function and divide it by 2. That means you have to integrate something squarey on the bottom.
• Nov 4th 2007, 03:45 AM
kalagota
pardon my young mind, but, aren't they the same? i am now confused.. or, maybe i have to check my notes.. Ü
• Nov 4th 2007, 03:50 AM
TheBrain
$\displaystyle 1/2 \int_0^{\frac{\pi}{2}} \frac {a^2}{(1+cos\theta)^2}d\theta$

<learns how to use the formula thing>

Well this is the integral that I end up with.
• Nov 4th 2007, 04:00 AM
kalagota
if the question was to find the area of that function $\displaystyle r=f(\theta) = \frac{a^2}{1+cos\theta}$ on the given interval, then it must be correct.. however, if you set-up the integral as it is, then i think, you'll just integrate it as it is..
• Nov 4th 2007, 04:02 AM
TheBrain
The actual question was

$\displaystyle r=f(\theta) = \frac{a}{1+cos\theta}$

integrated between the limits.
• Nov 4th 2007, 04:09 AM
kalagota
Quote:

Originally Posted by TheBrain
The actual question was

$\displaystyle r=f(\theta) = \frac{a}{1+cos\theta}$

integrated between the limits.

in my opinion, i'll just need to integrate it as it is (and not abuse the formula) since it didn't asked me to find the area of that curve from 0 to $\displaystyle \frac{\pi}{2}$.
• Nov 4th 2007, 10:42 AM
CaptainBlack
Quote:

Originally Posted by TheBrain
$\displaystyle 1/2 \int_0^{\frac{\pi}{2}} \frac {a^2}{(1+cos\theta)^2}d\theta$

<learns how to use the formula thing>

Well this is the integral that I end up with.

Why do you think this is anything to do with polars, $\displaystyle \theta$ in:

"Integrate (a^2) / (1 + cos theta) between 0 and pi/2 where a is just a constant."

is just a dummy variable of integration. The integral you want is:

$\displaystyle \int \frac{a^2}{1+\cos(\theta)}~d\theta$

Nothing more.

RonL
• Nov 4th 2007, 02:09 PM
TheBrain
I built the integral from the question and it is definately polar co-ordinates because it is from that chapter of the book and the integral limits are stated as being between angles pi/2 and 0. The integral that needs to be solved is definately

$\displaystyle 1/2 \int_0^{\frac{\pi}{2}} \frac {a^2}{(1+cos\theta)^2}d\theta$
• Nov 4th 2007, 02:39 PM
Plato
$\displaystyle \int {\frac{1}{{\left( {1 + \cos (x)} \right)^2 }} = \frac{1}{2}\tan \left( {\frac{x}{2}} \right) + \frac{1}{6}\tan ^3 \left( {\frac{x}{2}} \right)}$

Now work backwards to see how the antiderivative works.
BTW. Buy youcself a computer algebra system. There is one for @$40 if you have a PC. • Nov 4th 2007, 02:43 PM galactus You can let$\displaystyle {\theta}=2tan^{-1}(u), \;\ d{\theta}=\frac{2}{u^{2}+1}du, \;\ u=tan(\frac{\theta}{2})$After making the subs, we get:$\displaystyle \frac{a^{2}}{2}\int{\left[\frac{1}{2}u^{2}+\frac{1}{2}\right]}du\displaystyle \frac{a^{2}}{2}\left[\frac{1}{6}u^{3}+\frac{1}{2}u\right]$Now, resub and get:$\displaystyle \frac{a^{2}}{2}\left[\frac{1}{6}tan^{3}(\frac{\theta}{2})+\frac{1}{2}ta n(\frac{\theta}{2})\right]$Now, use the limits of integration and it boils down to$\displaystyle \frac{a^{2}}{3}$• Nov 4th 2007, 07:25 PM kalagota Quote: Originally Posted by TheBrain I built the integral from the question and it is definately polar co-ordinates because it is from that chapter of the book and the integral limits are stated as being between angles pi/2 and 0. The integral that needs to be solved is definately$\displaystyle
1/2 \int_0^{\frac{\pi}{2}} \frac {a^2}{(1+cos\theta)^2}d\theta
\$

will you qoute the whole question.. Ü