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Average area of image of rotated square

Suppose we have a unit square that is rotating in the x (or y) axis. At a point in time, it stops rotating. The image of it viewed from directly on top of it should be a rectangle. What is the average area of this rectangle?

Note by Daniel Liu
3 years, 10 months ago

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If it rotates an angle of \(\theta\) from the \(xy\)-plane, then the area (viewed from above) will be \(|\cos \theta|\).

Therefore, the average area is \(\dfrac{1}{2\pi}\displaystyle\int_{0}^{2\pi} |\cos \theta|\,d\theta = \dfrac{2}{\pi}\). Jimmy Kariznov · 3 years, 9 months ago

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How do you mean, it should be a rectangle? If you have a square, then you can rotate it all you want, but it stays a square, right? Tim Vermeulen · 3 years, 9 months ago

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@Tim Vermeulen But it is rotating in respect to the x axis, not the z axis. Daniel Liu · 3 years, 9 months ago

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Can you rephrase it? I don't get how it's rotating... Luca Bernardelli · 3 years, 9 months ago

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@Luca Bernardelli Hmm... I'm not really sure how to say it. Imagine a square. You stick a pole through two opposite midpoints of the square, and rotate it with that pole as the axis. That kind of rotating. Daniel Liu · 3 years, 9 months ago

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