Question

Use cylindrical shell method to find the volume of the solid obtained by rotating the region bounded by the given curves x^2=y and y^2=x about y=-1?

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Answer 1

`|V|=19/30pi`

Explanation:

Given:
first curve
`x^2=y`
second curve
`y^2=x`
Solving for x and y
`x=0, and x=1`
which are the lower and upper limits parallel to the axis `y=-1`

radius of the circle on first curve is given by
`r_1=y-(-1)`
`r_1=y+1`
For `" "y=x^2`
`r_1=x^2+1`

Area of circular section is
`A_1=pir_1^2`
`A_1=pi(x^2+1)^2`
`A_1=pi(x^4+2x^2+1)`

radius of the circle on second curve is given by
`r_2=y-(-1)`
`r_2=y+1`
For `" "y=sqrtx`
`r_2=sqrtx+1`

Area of circular section is
`A_2=pir_1^2`
`A_2=pi(sqrtx+1)^2`
`A_2=pi(x+2sqrtx+1)`

For `0<=x<=1," " x^2>sqrt`
Area of the annular portion is
`A=A_1-A_2`
`A=pi(x^4+2x^2+1)-pi(x+2sqrtx+1)`
`A=pi(x^4+2x^2+1)-(x+2sqrtx+1)`
`A=pi(x^4+2x^2+1-x-2sqrtx-1)`
`A=pi(x^4+2x^2-x-2sqrtx)`

The volume enclosed by the annular portion for distance dx is

`dV=Adx`

`dV=pi(x^4+2x^2-x-2sqrtx)dx`

Integrating between the limits x=0 and x=1

`intdV=intpi(x^4+2x^2-x-2x^(1/2))dx`

`V=pi(1/(4+1)x^(4+1)+2/(2+1)x^(2+1)-1/(1+1)x^(1+1)-2/(1/2+1)x^(1/2+1))_0^1`

`V=pi(1/5x^5+2/3x^3-1/2x^2-2/(3/2)x^(3/2))_0^1`

`V=pi(1/5x^5+2/3x^3-1/2x^2-4/3x^(3/2))_0^1`
Substituting the limits and taking the difference

`V=pi(1/5+2/3-1/2-4/3)`

`V=pi(6+20-15-40)/30`

`V=(26-55)/30pi`
`V=-19/30pi`
`|V|=19/30pi`