Answer:
The answer is "[tex]\frac{5\pi}{6}[/tex]"
Step-by-step explanation:
Please find the graph file.
[tex]h= y=2x-x^2\\\\r= x\\\\Area=2\pi\times r\times h\\\\= 2 \pi \times x \times (2x-x^2)\\\\= 2 \pi \times 2x^2-x^3\\\\volume \ V(x)=\int \ A(x)\ dx\\\\= \int^{x=1}_{x=0} 2\pi (2x^2-x^3)\ dx\\\\= 2\pi [(\frac{2x^3}{3}-\frac{x^4}{4})]^{1}_{0} \\\\= 2\pi [(\frac{2}{3}-\frac{1}{4})-(0-0)] \\\\= 2\pi \times \frac{5}{12}\\\\=\frac{5\pi}{6}\\\\[/tex]
