You have to use the "Cylinder Method" (aka "Shell Integration").
The volume of the solid obtained by rotating the region between f(x), y=0, x=a, and x=b is given by:
[tex] V=\displaystyle\int^b_a {2\pi x \left|f(x)\right|} \, dx [/tex]
In your case, [tex]f(x)=16-x^4[/tex]. So, substituiting f(x), a and b, and taking into account that in the interval (2,3) we can assume [tex]x^4\ \ge \ 16[/tex], and, therefore, [tex]\left|16-x^4\right| = -\left(16+x^4\right)[/tex], we get:
[tex]V=\displaystyle\int^3_2 {2\pi x\left|16-x^4\right|} \, dx = \int^3_2 {-2\pi x\left(16-x^4\right)} \, dx = -2\pi\int^3_2 {\left(16x-x^5\right)} \, dx [/tex]
We have made use of the fact that you can take constant factors out of the integral.
Then, being the integral of the sum equal to the sum of the integrals:
[tex]V = -2\pi\displaystyle\int^3_2 {\left(16x-x^5\right)} \, dx = -2\pi \int^3_2 {16x} \, dx + 2\pi \int^3_2 {x^5} \, dx[/tex]
For both integrals we'll use the power integration rule, and the Second Fundamental Theorem of Calculus:
[tex]\displaystyle \int {x^n} \, dx = \dfrac{x^{n+1}}{n+1}[/tex]
[tex]V= -\displaystyle 2\pi \int^3_2 {16x} \, dx + 2\pi \int^3_2 {x^5} \, dx = -2\pi\cdot16\cdot\left[\dfrac{x^2}{2}\right]^3_2 + 2\pi\cdot\left[\dfrac{x^6}{6}\right]^3_2[/tex]
[tex]V = -32\pi\left(\dfrac{3^2}{2}-\dfrac{2^2}{2}\right) + 2\pi\left(\dfrac{3^6}{6}-\dfrac{2^6}{6}\right) = \dfrac{425}{3}\pi[/tex]
