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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)

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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

  • 0
    What do you mean? I shaded my region.2011-12-13
  • 0
    What stuff did I forget to add?2011-12-13
  • 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 $$ \frac 12 \int_{-\pi/2}^{\pi/2} \, d\theta = \frac{\left( \frac {\pi}2 \right) - \left( -\frac{\pi}2 \right)}2 = \frac{\pi}2. $$2011-12-13
  • 0
    Do you know what's the point sets for you shaded area in terms of polar coordinates?2011-12-14
  • 0
    pi/2 and 3pi/2 should be symmetrically the same above and below2011-12-14
  • 0
    Was that $\frac{3\pi}2$ instead of $-\frac{3\pi}2$ before the edits?2011-12-14

2 Answers 2