Question

Solid formed after revolving the area between `y = x^2` and `y = x+2` around `y` axis?

Two functions
`y=x^2` and `y=x+2`
Around y axis

It will be `(16π)/3` because there is a solid inside other or we have to considerer solid 1 plus solid 2
`(5π)/6 + (16π)/3 = (37π)/6`

Answer 1

graph{(y-x^2)(y-x-2)=0 [-7.37, 10.41, -1.955, 6.934]}

Ignore the `x^2 le y le x+2` in Q2 as it will be within the solid created by revolving the part in Q1 about the y axis

That leaves:

`underbrace( pi int_0^4 x^2 \ bb(dy) )\_("Volume inside Parabola")- underbrace( pi r^2 h/3)_("Volume of Cone")`

With `y = x^2 qquad \ dy = 2 x dx`

`= 4pi int_0^2 x^3 \ bb(dx) - underbrace( pi * 2^2 * 2/3)_("Volume of Cone")`

`= pi [x^4 ]_0^2 - pi * 2^2 * 2/3`

`= 40/3 pi`