Rotate that function

Using the disc method , we know that: $V = \int_a^b \pi (f(x))^2 \, dx$

Here the function $f(x)=\sqrt x$ , thus the volume of revolution would be: $\begin{aligned} V &= \int_1^3 \pi(\sqrt x)^2 \, dx\\ &= \int_1^3 \pi x \, dx\\ &= \left[\pi \dfrac{x^2}2\right]^3_1= \dfrac{9\pi}2-\dfrac{\pi}2 \\ &= 4\pi\\ \end{aligned}$