A broken scale shows a false weight that is always less than the actual weight by the same number of ounces. When a box is put o
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The disk method will only involve a single integral. I've attached a sketch of the bounded region (in red) and one such disk made by revolving it around the y-axis.
Such a disk has radius x = 1/y and height/thickness ∆y, so that the volume of one such disk is
π (radius) (height) = π (1/y)² ∆y = π/y² ∆y
and the volume of a stack of n such disks is
where is a point sampled from the interval [1, 5].
As we refine the solid by adding increasingly more, increasingly thinner disks, so that ∆y converges to 0, the sum converges to a definite integral that gives the exact volume V,
