First solve x³=9x
[tex]x^3-9x=0
\\x(x^2-9)=0
\\x(x-3)(x+3)=0
\\x=0, x=3,x=-3[/tex]
We can ignore x = -3 because it is not in the first quadrant.
So our integral is going to go from x=0 to x=3.
Now we can use the formula
[tex]V=\int_{a}^{b}\pi f(x)^2dx[/tex]
[tex]\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx
\\\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx
\\\pi\int\limits_{0}^{3}(81x^2-x^6)dx
\\2\times \int\limits_{0}^{3}[\pi(9x)^2-\pi(x^3)^2]dx
\\\pi\left[81\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{3} \\ \\\frac{2916\pi}{7}[/tex]
