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In here, question 6, they calculate the volume of a torus by using cross-sections across the x-axis.
However, I have found an easier way to do it, using the same diagram: $$\int_0^{2\pi R} \pi r^2 \, dR = 2\pi^2 Rr^2.$$ (I skipped a few steps because I couldn't use begin{align} and end{align}.)

Why did they do it calculating cross-sections instead of using this easier method?

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    Yeah, I did it the same way. I just unraveled the donut into a dough cylinder, and just found the volume of that. – 2017-01-31
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    It would be the base (πr^2) times the height (2πR) = (2πRr^2) – 2017-01-31
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    @SakethMalyala Intuitive as that may be, the problem with that is that you must *prove* that the volume does not change when you *unravel*. – 2017-01-31
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    @SakethMalyala Please look [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) to format your equations. – 2017-01-31

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They wanted to use a more rigorous method, as using pure geometry will not always work.