Question

How do I find the volume of the solid generated by revolving the region bounded by `y=x^2`, `y=0`, and `x=2` about the `x`-axis? The `y`-axis?

Answer 1

1)`Volume=piint_0^2x^4*dx=(32/5)pi (unite)^3`

2)`Volume=piint_0^4[(2^2)_2-(sqrty^2)_1]*dy=piint_0^4[4-y]*dy=8pi (unite)^3`

Explanation:

the rose region is revolving about the x-axis and y-axis

1)when the shaded region revolving a bout x-axis

`Volume=piint_a^by^2*dx`

`Volume=piint_0^2y^2*dx=piint_0^2x^4*dx=pi[1/5*x^5]_0^2`

`=pi[(32/5)-0]=(32/5)pi (unite)^3`

2)when the shaded region revolving about the y-axis

`Volume=piint_d^c[(x^2)_2-(x^2)_1]*dy`

`Volume=piint_0^4[(2^2)_2-(sqrty^2)_1]*dy`

`=piint_0^4[4-y]*dy=pi[4y-1/2y^2]_0^4`

`=pi[16-8]=8pi (unite)^3`