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What is the first moment of an isosceles right triangle about the vertex of the right angle?

I was thinking to look at it as

$$\int_{-\pi/4}^{\pi/4} \int_{0}^{\sec(\theta)} r^2drd\theta.$$

Does that make any sense?

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  • $\begingroup$ Can you draw a picture? (having a hard time visualizing it) $\endgroup$ – apnorton Mar 3 '13 at 20:13
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Your integral is correct (for an isosceles right triangle with unit altitude and hence unit area), and it seems like a good way to calculate the first moment, since calculating it in Cartesian coordinates is also complicated and involves two non-trivial integrations instead of one.

Wolfram|Alpha finds a nice form for the result:

$$\displaystyle\frac13\left(\sqrt2+2\tanh^{-1}\tan\frac\pi8\right)=\frac13\left(\sqrt2+2\tanh^{-1}\left(\sqrt2-1\right)\right)\;.$$

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