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
![\displaystyle V_n = \sum_(i=1)^n \pi {y_i}^2 \Delta y](https://img.qammunity.org/2023/formulas/mathematics/high-school/disnr0rt0i1s1u6upmto6b7nm9nng8xof0.png)
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,
![\displaystyle V = \lim_(n\to\infty) V_n = \int_1^5 (\pi)/(y^2) \, dy](https://img.qammunity.org/2023/formulas/mathematics/high-school/nk9vpcumophnwf7ipjw4g0ed70birso7yb.png)
![V = -\frac\pi y\bigg|_(y=1)^(y=5) = \boxed{\frac{4\pi}5}](https://img.qammunity.org/2023/formulas/mathematics/high-school/64ay1hmk4y9zknnko42ctxrb78ws5egej2.png)