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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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    @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.2012-07-02

2 Answers 2

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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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    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.2012-07-03
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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.$