Question

How to find the volume of a solid that is enclosed by y=x^(2) and y=x^(1/2), and revolved around x=-6?

I know that we need to use the washer method and do the integration in terms of y and that it needs to be integrated between 0 and 1 but no matter what I try I can't seem to get the right answer

Answer 1

`V=43/10pi`

Explanation:

The axis of rotation which is common has been given by
`x=-6`

The intersection of the curves `y=x^2, and y=x^(1/2)` form the lower limit and upper limit of the integration

Equating rhs of the curves

`x^2=x^(1/2)`

`x^2-x^(1/2)=0`

`x^(1/2)(x^(3/2)-1)=0`

`x^(1/2)=0`, implying `x=0`, where
`y=x^2=0^2=0`

`x^(3/2)-1=0`, implying
`x^(3/2)=1`

Squaring both sides

`x^3=1`

Among the three roots, two are imaginary/

The real root is `x=1`, where
`y=x^2=1^2=1`

The points on the curves are rotating at a radius of curvature given by perpendicular distance from the point on the curve to the axis.

The upper curve y=x^2 and the lower curve y=x^1/2 form the annular ring of radii r1 and r2

`r=y+6`

Area is given by

`A_1=pir_1^2=pi(y+6)^2=pi(x^2+6)^2`

`A_1=pi(x^4+2xxx^2xx6+6^2)=pi(x^4+12x^2+36)`

`A_2=pir_2^2=pi(y+6)^2=pi(x^(1/2)+6)^2`

`A_2=pi(x+2xxx^(1/2)xx6+6^2)=pi(x+12x^(1/2)+36)`

The area of the collar A is given by A1-A2

`A=A_1-A_2`

`A=pi(x^4+12x^2+36)-pi(x+12x^(1/2)+36)`

`A=pi(x^4+12x^2-x-12x^(1/2))`

Volume swept is obtained by integrating the above expression betwen the limits x=0, and x=1

`dV=|int_0^1Adx|`

`=|int_0^1pi(x^4+12x^2-x-12x^(1/2))dx|`

`=|pi(1/5x^5+12/3x^3-1/2x^2-12xx2/3x^(3/2))_0^1|`

`=|pi(1/5+4-1/2-8)|`

`=|pi(2/10-5/10+4-8)|`

`=|pi(-3/10-4|)`

`=|pi(-3/10-4/10)|`

`=|pi(-43/10)|`

`V=43/10pi`