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21. find the volume of the solid of revolution formed if the area enclosed between the curves y=x² and y=(x-2)² is rotated about the x-axis using integration

Sagot :

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.
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