Method of Cylindrical Shells

In addition to the slicing methods, there is another way to calculate the volume of a solid of revolution. While the slicing method is based on the idea of approximating cross sections taken perpendicular to the axis of rotation, the method of cylindrical shells uses approximating hollow cylinders centered about the axis of rotation.

To describe this method, we let $R$ be the region bounded by the graph of a continuous nonnegative function $f$ and the $x$-axis for MATH . If the region $R$ is rotated about the $y$-axis, a solid $S$ is generated.

To calculate the volume $V$ of $S$, we partition the interval $[a,b]$ using the equal-length partition
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where MATH Then we choose the numbers $t_{j}$ to be the midpoints of the subintervals. That is, MATH Using these choices, we approximate the region $R$ by rectangles with base $[x_{j-1},x_{j}]$ and height $f(t_{j})$ . If we rotate one of these rectangles about the $y$-axis, we obtain a cylindrical shell. Combining all such shells, we obtain a solid whose volume approximates the volume of $S.$

Consider the jth shell. Its height is $f(t_{j}),$ and the area of its base is MATH Consequently, its volume $V_{j}$ is
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Thus, by summing the volumes of all of the shells, we approximate the volume $V$ by
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To see that the right-hand side of this approximation is a Riemann sum, we factor
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and rewrite the approximation as
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Since MATH we have
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and taking the limit as MATH these Riemann sums approach
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Assuming that these sums also limit to $V,$ we obtain the following volume formula for $S.$

If the region $R$ bounded by the graph of the continuous nonnegative function $f$ and the $x$-axis, for MATH , is rotated about the $y$-axis, the volume $V$ of the resulting solid is
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EXAMPLE 1

The region bounded by the graph of $y=-2x^{2}+8x-6$ and the $x$-axis is rotated about the $y$-axis. Find the volume $V$ of the resulting solid.

Solution

Since MATH the region lies between the lines $x=1$ and $x=3.$
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Equation
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can be generalized to regions bounded below by curves other than the $x$-axis, as the following example shows.

EXAMPLE 2

The region $R$ is bounded by the graphs of $f(x)=-2x^{2}+8x-6$ and $g(x)=2x-6 $ . Find the volume of the solid generated by revolving $R$ about the $y$-axis.

Solution

To find the points of intersection, we set $f(x)=g(x)$ and obtain the equation
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or
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so $x=0$ or $x=3.$ When the interval $[0,3]$ is partitioned, the approximating rectangles are bounded above by $f(x)=-2x^{2}+8x-6$ and below by $g(x)=2x-6.$ Since the factor $f(x)$ represents the height of the approximating rectangles, we modify equation to the following:
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REMARK We may state the generalization of equation
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observed in Example 2 as follows: If the region $R$ bounded above by the graph of $y=f(x)$ and below by the graph of $y=g(x),$ for MATH is rotated about the $y$-axis, the volume of the resulting solid is
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EXAMPLE 3

Find the volume of the solid obtained by revolving about the line $x=-1$ the region $R$ bounded by the graphs of $f(x)=x$ and $g(x)=(x-2)^{2}.$

Solution The region $R$ that is bounded above by the graph of $f(x)=x$ and below by the graph of MATH To find the points of intersection of these two graphs we set $f(x)=g(x),$ which gives the equation
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or


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The points of intersection are therefore $(1,1)$ and $(4,4).$

A vertical line segment through this region generates a circular band of height
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Since the axis of rotation is the vertical line $x=-1,$ the radius of this circular band is $x-(-1)=x+1.$ We must therefore replace the factor $x$ in the integrand in formula
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by the factor $x+1.$ The volume is therefore
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