First factor -12m^n - 49mn - 44n^2 to get -(4n+3m)(11n + 4m) then the equation would be:
-(3m + 4n)(4m + 11n) / (-3m - 4n)
Then, cancel out the like terms and the final answer would be
4m + 11n
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://tex.z-dn.net/?f=%5Cdisplaystyle%20V_n%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20%5Cpi%20%7By_i%7D%5E2%20%5CDelta%20y)
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 \frac{\pi}{y^2} \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%20%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20V_n%20%3D%20%5Cint_1%5E5%20%5Cfrac%7B%5Cpi%7D%7By%5E2%7D%20%5C%2C%20dy)
![V = -\dfrac\pi y\bigg|_{y=1}^{y=5} = \boxed{\dfrac{4\pi}5}](https://tex.z-dn.net/?f=V%20%3D%20-%5Cdfrac%5Cpi%20y%5Cbigg%7C_%7By%3D1%7D%5E%7By%3D5%7D%20%3D%20%5Cboxed%7B%5Cdfrac%7B4%5Cpi%7D5%7D)