Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $y=0$, $x=1$, and $x=2$ about the line $x=4$.
Here is what I attempted with the shell method, but it is clearly wrong:
1 Answer
$\begingroup$If we are rotating about the $y$-axis ($x=0$), then the formula is
$$2\pi\int_a^b xf(x) \, dx.$$
For your problem, rather than $x=0$, the rotation is about $x=4$.
The integral of interest should be
$$2\pi \int_1^2 (4-x)x^2\, dx.$$
Alternatively, to some, it might be easier to view it as rotating the region bounded by the curves $y=(x-8)^2, y=0, x=6,$ and $x=7$ about the line $x=4$. This can be observed by a reflection.
which is equivalent to rotating the region bounded by the curves $y=(x-4)^2, y=0, x=2,$ and $x=3$ about the line $x=0$. This can be observed by a translation.
That is you can also just evaluate $$2\pi \int_2^3x(x-4)^2 \, dx.$$
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