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
[tex]\displaystyle V_n = \sum_{i=1}^n \pi {y_i}^2 \Delta y[/tex]
where [tex]y_i[/tex] 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,
[tex]\displaystyle V = \lim_{n\to\infty} V_n = \int_1^5 \frac{\pi}{y^2} \, dy[/tex]
[tex]V = -\dfrac\pi y\bigg|_{y=1}^{y=5} = \boxed{\dfrac{4\pi}5}[/tex]
