Question

Find the volume of the solid of revolution formed by revolving the region bounded by f(x)=-3x^2+8 and g(x)=3x^2+2 about the y-axis. A) sketch the graph of the region bounded by graphs. B) write the integral to compute the volume of the solid formed. C) calculate the volume of the solid.

Answer 1

`3\pi` (cubic units)

Explanation:

Shell Method (Vertical Axis)

`\displaystyle A=2\pi\int^b_ar(x)h(x)\,dx`

Radius:
`r(x)=x`

Height:
`h(x)=(-3x^2+8)-(3x^2+2)=-6x^2+6`

Bounds:
`[a,b]=[0,1]` (since rotation is about the y-axis)

Set up integral and evaluate

`\displaystyle A=2\pi\int^1_0x(-6x^2+6)\,dx\\\\A=2\pi\int^1_0(-6x^3+6x)\,dx\\\\A=2\pi\biggr(-(3)/(2)x^4+3x^2\biggr)\biggr|^1_0\\\\A=2\pi\biggr(-(3)/(2)(1)^4+3(1)^2\biggr)\\\\A=2\pi\biggr(-(3)/(2)+3\biggr)\\\\A=2\pi\biggr((3)/(2)\biggr)\\\\A=3\pi`

Attached is a graph with the shaded region, so I hope it helps!