# Area computed using polar coordinates

I have $r = \sqrt{\theta}$

http://www.wolframalpha.com/input/?i=cartesian+r+%3D+%5Csqrt%7Btheta%7D+

The graph given in the book ends at the first time it approaches to right side of the x axis (or 2pi). I attempted to set up an integral that cut each section of the graph so I have 4 section to compute that integral in and it did not even give close to the correct answer ($\pi^2$).

I tried to do

$$\int_0^ {\pi/2} \sqrt {\theta} d \theta$$ $$\int_{\pi/2}^ {\pi} \sqrt {\theta} d \theta$$ $$\int_\pi^ {3\pi/2} \sqrt {\theta} d \theta$$ $$\int_{3\pi/2}^ {2\pi} \sqrt {\theta} d \theta$$

This basically gives me just the integral of $\frac{1}{4} \theta^{2}$ evaluated $2\pi$ since everything else cancels out.

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And the integrals you attempted to compute and your subsequent computations were... ? – Arturo Magidin Jul 2 '12 at 22:01
The area in polar coordinates is not computed as $\int f(\theta)\,d\theta$. The formula is $\frac{1}{2}\int (f(\theta))^2\,d\theta$. Wrong formula, wrong answer. – Arturo Magidin Jul 2 '12 at 22:08
@ArturoMagidin The formula isn't the problem. I have tried it with the correct formula, I just typed it up wrong. – user138246 Jul 2 '12 at 22:09
And, of course, I'm supposed to just know that you used the right formula but typed the wrong one... I think part of the problem might be carelessness. – Arturo Magidin Jul 2 '12 at 22:15
@ArturoMagidin I didn't realize I typed it out wrong I was just trying to type it out quickly to exit this room that is 110 degrees as fast as possible. – user138246 Jul 2 '12 at 22:30

You have no need to divide the integral from $0$ to $2\pi$ into a sum of other integrals.

$$r=f(\theta)=\sqrt{\theta}$$

The integral from $0$ to $2\pi$ of $f(\theta)$ is given by:

$$\frac{1}{2}\int_{a}^{b}(f(\theta))^2\, d\theta=\frac{1}{2}\int_{0}^{2\pi}\sqrt{\theta}^2\, d\theta=\frac{1}{2}\int_{0}^{2\pi}\theta\, d\theta=\frac{1}{2}\left[\frac{1}{2}\theta^2\right]_0^{2\pi}=\frac{1}{4}((2\pi)^2-0^2)=\frac{4\pi^2}{4}=\pi^2$$

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Why do I not have to divide it up? Now it is going above and below the axis. – user138246 Jul 2 '12 at 22:13
@Jordan Ah, but it is not a regular function! It is in polar coordinates. – Argon Jul 2 '12 at 22:16
That doesn't make any sense to me I guess...if I were to graph this in cartesian then it would all be in one quadrant? – user138246 Jul 2 '12 at 22:17
Think about it as absolute area. To test it out, try integrating $f(\theta)=R$ from $0$ to $2\pi$ (a circle of radius $R$) and see if it equals $\pi R^2$ – Argon Jul 2 '12 at 22:19
You may also like to think about this from the derivation of the integral. Regular integrals can be thought as Reimann sums, i.e. rectangles with base of $\Delta x$ and height of $f(x_n)$ at $x_n$. It makes sense then to have "negative" area. However, with polar coordinates, think of it as a sum of triangles all touching the origin. You can see that negative area does not make sense. It is also good to remember that in the integral, $f(\theta)$ is squared, and thus always positive. – Argon Jul 3 '12 at 1:15

We know that $$A = \frac{1}{2}\int_0^{2\pi}r^2d\theta = \frac{1}{2}\int_0^{2\pi}\theta d\theta = \left[\frac{1}{4}\theta^2\right]_{0}^{2\pi} = \pi^2.$$

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