Polar coordinates and multivariable integrals.

[Minihoudini]

Homework Statement

Im righting this down for my roommates since he's having tons of trouble trying to figure this out and I can't answer it.
also sorry for having to hotlink it.
http://i.imgur.com/afShz.jpg

the equation is on the image since its very difficult to type it all out. Apologies about that.
we understand everything up till the 2 is introduced and the integral goes from 0 to pi/6. We don't understand how you are able to just introduce a 2 and take out the -pi/6.

 

[cepheid]

This combines a simple property of integrals along with the fact that the function of theta is an even function (meaning that it is symmetric around the y-axis).

The property of integrals is that you can split the interval of integration up into a bunch of sub-intervals. The integral over the overall interval is then equal to the sum of the integrals over the subintervals. For instance, you can split the interval from -pi/6 to +pi/6 up into two sub-intervals. The first interval goes from -pi/6 to 0, and the second interval goes from 0 to +pi/6. According to the rule I just stated, then, we can write:\int_{-\pi/6}^{\pi/6} f(\theta)\,d\theta = \int_{-\pi/6}^{0} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\thetaNow, if f(θ) is an even function, then it is symmetric around the y-axis. This means that the area under the curve from -pi/6 to 0 is exactly the same as the area under the curve from 0 to pi/6. Therefore, you can substitute one integral for the other (they are numerically equivalent):\int_{-\pi/6}^{\pi/6} f(\theta)\,d\theta = \int_{-\pi/6}^{0} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\theta = \int_{0}^{\pi/6} f(\theta)\,d\theta + \int_{0}^{\pi/6} f(\theta)\,d\thetaIn the last expression, I just replaced the integral from -pi/6 to 0 with an integral from 0 to pi/6, since they are the same. Obviously the last sum can then by written as= 2\int_0^{\pi/6} f(\theta)\,d\theta

 

[Minihoudini]

thank you, this makes a lot more sense.