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Shell integration (the shell method in integral calculus) is a method for calculating the
volume Volume is a measure of regions in three-dimensional space. It is often quantified numerically using SI derived units (such as the cubic metre and litre) or by various imperial or US customary units (such as the gallon, quart, cubic inch) ...
of a solid of revolution, when integrating along an axis ''perpendicular to'' the axis of revolution. This is in contrast to disc integration which integrates along the axis ''parallel'' to the axis of revolution.


Definition

The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the -plane around the -axis. Suppose the cross-section is defined by the graph of the positive function on the interval . Then the formula for the volume will be: :2 \pi \int_a^b x f(x)\, dx If the function is of the coordinate and the axis of rotation is the -axis then the formula becomes: :2 \pi \int_a^b y f(y)\, dy If the function is rotating around the line then the formula becomes: :\begin \displaystyle 2 \pi \int_a^b (x-h) f(x)\,dx, & \text\ h \le a < b\\ \displaystyle 2 \pi \int_a^b (h-x) f(x)\,dx, & \text\ a < b \le h, \end and for rotations around it becomes :\begin \displaystyle 2 \pi \int_a^b (y-k) f(y)\,dy, & \text\ k \le a < b\\ \displaystyle 2 \pi \int_a^b (k-y) f(y)\,dy, & \text\ a < b \le k. \end The formula is derived by computing the double integral in
polar coordinates In mathematics, the polar coordinate system specifies a given point (mathematics), point in a plane (mathematics), plane by using a distance and an angle as its two coordinate system, coordinates. These are *the point's distance from a reference ...
.


Derivation of the formula


Example

Consider the volume, depicted below, whose cross section on the interval , 2is defined by: :y = 8(x-1)^2(x-2)^2 With the shell method we simply use the following formula: :V = 16 \pi \int_1^2 x ((x-1)^2(x-2)^2) \,dx By expanding the polynomial, the integration is easily done giving \pi cubic units.


Comparison With Disc Integration

Much more work is needed to find the volume if we use disc integration. First, we would need to solve y = 8(x-1)^2(x-2)^2 for . Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow. After integrating each of these two functions, we would subtract them to yield the desired volume.


See also

* Solid of revolution * Disc integration


References

* * Frank Ayres, Elliott Mendelson. '' Schaum's Outlines: Calculus''. McGraw-Hill Professional 2008, . pp. 244–248 () {{Calculus topics Integral calculus