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Find the volume of the solid obtained by rotating the region bounded by the curves y=x^6, y=1, about the line y=6
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This is the situation: (see image) We can simplify a bit by moving everything 6 units down and rotate about the x-axis. If we use the cylindrical shell method, we get the integral\[V= \int\limits_{-1}^{1}\pi(f(x) - 6)^2dx=2\pi \int\limits_{0}^{1}(x^6-6)^2dx\]I saved on the calculations (and errors!) by using the symmetry of the thing: the integral from -1 to 1 is equal to twice the integral from 0 to 1. Because \(\pi\) is a constant, it can also be put in front of the integral. Now all you have to do is expand the brackets and integrate every term.
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