What you need to do is get hold of the area underneath the curve y=x² from x=1 to x=0. You then spin this area 360 degrees about the x-axis and double the result as there is symmetry between y=x² and y=(x-2)².
Use the formula:
[tex]Volume=\int _{ a }^{ b }{ \pi { y }^{ 2 } } dx[/tex]
Ok, so let's solve the problem...
[tex]V=2\int _{ 0 }^{ 1 }{ \pi { x }^{ 4 } } dx\\ \\ =2{ \left[ \frac { \pi { x }^{ 4+1 } }{ 4+1 } \right] }_{ 0 }^{ 1 }[/tex]
[tex]\\ \\ =2{ \left[ \frac { \pi { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }\\ \\ =2\left\{ \left( \frac { \pi }{ 5 } \right) -\left( 0 \right) \right\} \\ \\ =\frac { 2 }{ 5 } \pi [/tex]
Answer:
[tex]\frac { 2 }{ 5 } \pi [/tex] units cubed.
