Answer
$\dfrac{ 72}{5}$
Work Step by Step
We need to integrate the integral as shown below:
$V= \int_{0}^{6} y^2 dx= \int_{0}^{6} [36-24 \sqrt {6x} +36 x-4 \sqrt 6 x^{3/2} +x^2] dx \\= [36x-16 \sqrt {6}x^{3/2} +18x^2 -\dfrac{8 \sqrt 6 x^{5/2}}{5} + \dfrac{x^3}{3}]_0^6 \\=\dfrac{ 72}{5}$
