I assume the first equation is [tex]y=\frac1{x^2}[/tex]. Using the method of washers, the volume is
[tex]\displaystyle\pi\int_1^8\left(\frac1{x^2}+3\right)^2-3^2\,\mathrm dx=\pi\int_1^8\frac1{x^4}+\frac6{x^2}\,\mathrm dx=\boxed{\frac{8575\pi}{1536}}[/tex]
