Answer:
Step-by-step explanation:
[tex]z=x,\quad z=x^2,\quad 0\leq y\leq 5\\x=x^2,\quad x-x^2=0,\quad x(1-x)=0,\quad x=0\vee x=1\\V=\int_0^1 (x-x^2)dx\cdot\int_0^5dy=\left.\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\right|_0^1\cdot 5=\left(\frac{1}{2}-\frac{1}{3}\right)\cdot 5=\frac{5}{6}[/tex]
