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# 1.3: Volume by Cylindrical Shells

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## Cylindrical Shells

Consider rotating the region between the curve

$y = x^2, \nonumber$

the line

$x = 2, \nonumber$

and the x-axis about the y-axis.

If instead of taking a cross section perpendicular to the y-axis, we take a cross section perpendicular to the x-axis, and revolve it about the y-axis, we get a cylinder. Recall that the area of a cylinder is given by:

$A = 2\pi r h \nonumber$

where $$r$$ is the radius of the cylinder and $$h$$ is the height of the cylinder. We can see that the radius is the x coordinate of the point on the curve, and the height is the y coordinate of the curve. Hence

$A(x) = 2\pi xy = 2\pi x(x^2). \nonumber$

Therefore the volume is given by

\begin{align*} \text{Volume} &= 2\pi \int_{0}^{1} x^3 dx \\ &= \dfrac{\pi}{2}. \end{align*}

Example 1

Find the volume of revolution of the region bounded by the curves

• $$y = x^2 + 2$$
• $$y = x + 4$$
• and the y-axis

about the y-axis.

Solution

We draw the picture with a cross section perpendicular to the x-axis.

The radius of the cylinder is $$x$$ and the height is the difference of the $$y$$ coordinates:

$h = (x + 4) - (x^2 + 2). \nonumber$

We solve for $$b$$.

\begin{align*} (x+4)&=(x^2+2) \\ x^2-x-2&=0 \\ (x-2)(x+1)&=0 \end{align*}

So that $$b = 2$$. Hence the volume is equal to

\begin{align*} 2 \pi \int_{0}^{2} \big[ (x+4) -(x^2+2) \big] dx &= 2 \pi \int_{0}^{2}(x^2+2x-x^3) dx \\ &= 2 \pi \left(\dfrac{x^3}{3}+x^2-\dfrac{x^4}{4} \right]_{0}^{2} \\ &= 2 \pi \Big( \dfrac{8}{3}+4 -4 \Big) \\ &= \dfrac{16 \pi}{3}. \end{align*}

## Exercises

1. $$y=x^2-3x+2$$, $$y=0$$ about the y-axis
2. $$y=x^2-7x+6$$, $$y=0$$ about the y-axis
3. $$x=1-y^2$$, $$x=0$$ (first quadrant) about the x-axis
4. $$y=x\sqrt{1+x^3}$$, $$y=0$$, $$x=2$$ about the y-axis
5. $$(x-1)^2+y^2=1$$ about the y-axis
6. $$x^2+(y-1)^2=1$$ about the x-axis
7. $$y=x^2-2x+1$$, $$y=1$$ about the line $$x=3$$

Answers