Volume of a solid of revolution (quadratic and linear boundary) Sketch and find the volume of a solid bounded by the curve $y = x^{2} + 2$ and the line $y = 4 - x$ rotated about $y = 6.$
Why is the formula for this not $$\pi \int\limits_a^b ([f(x)]^{2} - [g(x)]^2)\,dx ?$$
 A: The bounded region is given below.

Put a vertical strip inside the region. If we rotate it to the line $y=6$, a washer is obtained and its volmue is given by
$$dV=\pi([6-(x^2+2)]^2)-[6-(4-x)]^2)$$ and so the required volume is given by
$$V=\int_{-2}^1 dV=\dots$$
and its your turn to do the computations. Hope this help.
A: Perhaps this figure will help:

So you can find the volume by integrating the truncated (yellow) ovoid and subtracting the (red) cone.
Ovoid
$V_o = \int\limits_{x=-2}^1 \pi r^2 dx = \int\limits_{x=-2}^1 \pi (x^2 + 2 -6)^2 dx = {153 \pi \over 5}$
Cone
$V_c = {1 \over 3} \pi r^2 h$ where here $r = 2$ and $h = 3$, so $V_c =4 \pi$
So $V = V_o - V_c = {153 \pi \over 5} - 4 \pi = {133 \pi \over 5}$.

As to your question, the solution is
$\pi \int f^2(x) - g^2(x)\ dx$ where you have to be careful about $f(x)$ and $g(x)$.  Here $f(x) = x^2 + 2 - 6$ and $g(x) = 4 - x - 6$.

If anyone wants to plot such a figure in Mathematica, here is my code:
myovoid = 
  RegionPlot3D[(x + 2) < Sqrt[(y - 6)^2 + z^2] < (4 - x^2),
  {x, -5, 5}, {y, 0, 10}, {z, -5, 5},
  PlotPoints -> 100,
  AxesLabel -> {Text[Style["x", 20]], Text[Style["y", 20]], 
    Text[Style["z", 20]]},
  PlotStyle -> Opacity[0.5],
  AxesOrigin -> {0, 0, 0},
  MeshFunctions -> Function[{x}, x],
  Mesh -> 5];

mycone =
RegionPlot3D[(x + 2) > Sqrt[(y - 6)^2 + z^2 ] && x < 1,
 {x, -5, 1}, {y, 0, 10}, {z, -5, 5},
 PlotPoints -> 100,
 AxesLabel -> {Text[Style["x", 20]], Text[Style["y", 20]], 
   Text[Style["z", 20]]},
 PlotStyle -> {Red, Opacity[0.1]},
 AxesOrigin -> {0, 0, 0},
 MeshFunctions -> Function[{x}, x],
 Mesh -> 5];

Show[myovoid, mycone,
 Graphics3D[{Red, Thickness[0.01], Line[{{-4, 6, 0}, {4, 6, 0}}]}]]

