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Consider this problem.

Suppose I just want to find the area of a unit circle in polar coordinates. So let $r = 1$. Now I want to (for sake of theory of this question) integrate this over the shaded region (see picture)

enter image description here

Now I know that $$\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta = \pi$$ Because I am integrating over $\pi/2$ from below and to above.

But why is it that if I do $$\frac{1}{2}\int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}} d\theta = -\pi$$ I thought that integrating from $\frac{-\pi}{2}$ is the same as integrating $\frac{3\pi}{2}$

Thank you for reading

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What do you mean? I shaded my region. –  Jake Dec 13 '11 at 23:46
    
What stuff did I forget to add? –  Jake Dec 13 '11 at 23:47
5  
You didn't forget how to integrate, you forgot how to add stuff. $$ \int_{-3\pi/2}^{\pi/2} d \theta = \frac {\pi}2 - \left( \frac{-3\pi}2 \right) = 2\pi \neq -2\pi. $$ Also note that the first integral is wrong, because $$ \frac 12 \int_{-\pi/2}^{\pi/2} \, d\theta = \frac{\left( \frac {\pi}2 \right) - \left( -\frac{\pi}2 \right)}2 = \frac{\pi}2. $$ –  Patrick Da Silva Dec 13 '11 at 23:47
    
Do you know what's the point sets for you shaded area in terms of polar coordinates? –  Jack Dec 14 '11 at 0:01
    
pi/2 and 3pi/2 should be symmetrically the same above and below –  Jake Dec 14 '11 at 0:57

2 Answers 2

From 4:00 AM to 5:00 AM is just one hour, but from 4:00 AM to 5:00 PM is 13 hours, even though 5:00 is the same point on the circular clock face both times. If the only concern were position of the clock's hands, then they're the same thing. And if the only concern is the value of sine and cosine---thus the location on the circle, then $-\pi/2$ is the same as $3\pi/2$. But the arc length from $0$ to $3\pi/2$ in the positive direction is not at all the same as that from $0$ to $-\pi/2$ in the negative direction.

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First, the area of a circle is $\pi r^2$, or $\pi$ for the unit circle. Thus half the area would be $\pi/2$. This corresponds to your first integral, as, $$\frac{1}{2}\int_{-\pi/2}^{\pi/2} \; d\theta = \frac{1}{2}\left[\frac{\pi}{2}+\frac{\pi}{2}\right] = \frac{\pi}{2}.$$ For the full area, $$\frac{1}{2}\int_{-3\pi/2}^{\pi/2} \; d\theta = \frac{1}{2}\left[\frac{\pi}{2}+\frac{3\pi}{2}\right] = \pi.$$

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