Respuesta :

[tex]\begin{gathered} y=x^2 \\ x=0 \\ y=4 \\ 4=x^2 \\ \sqrt{4}=x \\ x=2 \\ V=\int A(x)dx \\ \\ A(x)=\pi r^2 \\ A(x)=\pi(x^2)^2=\pi x^4 \\ \\ V=\int_0^2\pi x^4dx=\frac{\pi x^5}{5}|^2_0=\frac{\pi(2)^5}{5}\text{ -0=}\frac{32\pi}{5} \\ The\text{ volume is }\frac{32\pi}{5} \end{gathered}[/tex]