Subject: Calculus

# Shell Integration

Some volume problems are very difficult to handle by the disk integration. For instance, let�s consider the problem of finding the volume of the solid obtained by rotating about the -axis the region bounded by y=2x^2- x^3 and y=0 as shown in the figure below. We could slice perpendicular to the y-axis to get a washer but it would be very difficult to solve for the consequent problem. By using the washer method, we are to compute the inner radius and the outer radius of the washer and would need to solve the cubic equation for y=2x^2- x^3 and y=0.

In order to solve this type of problem, we introduce a new method called the shell integration or more commonly known as the method of cylindrical shells. Take for example the Figure shown below which shows a cylindrical shell with an inner radius r_1, outer radius r_2, and height h. Its volume V is calculated by subtracting the volume V_1 of the inner cylinder from the volume V_2 of the outer cylinder:

V = V_2 - V_1=\pi r_2^2 h- \pi r_1^2h=\pi (r_2^2 - r_1^2)h
=\pi (r_2-r_1)(r_2+r_1)h=2 \pi \frac{\left (r_2+r_1\right )}{2}(r_2-r_1)h

If we let \delta r be equal to (r_2 - r_1) and r = \frac{r_2 + r_1 }{2}, we can then rewrite the volume of the cylindrical shell to

V=2 \pi r \delta r h

which can be remembered as

V=(Circumference)(thickness)(height)

Now let us consider the figure shown below, This is a solid obtained by rotating the figure on the left side about the y-axis and is bounded by y = f(x) where f(x) \geq 0, y = 0, x = a, and x =b where b > a \geq 0. To solve for the volume of this solid, we divide the interval [a, b] into n subintervals \left [x_{i-1}, x_i\right ] of equal width \delta x and let x _i be the midpoint of the i^th subinterval. If the rectangle with base \left [x_{i-1}, x_i\right ] and height f(\bar{x}_i) is rotated about the y-axis, then the result is a cylindrical shell with average radius \bar{x}_i, height f(\bar{x}_i), and thickness \delta x which is best illustrated by the figures below. Therefore, using the formula of the volume derived above, we have

V=2\pi \bar{x}_{i}\delta x \left [ f(\bar{x}_{i}) \right ]

Therefore an approximation to the volume V of the solid is given by the sum of the volumes of these shells which is

V\approx \sum_{i=1}^{n}V_{i}=\sum_{i=1}^{n}2\pi \bar{x}_{i}\delta x \left [ f(\bar{x}_{i}) \right ]

This shows that the approximation becomes better as n \to \infty. In terms of integral we can rewrite this as

\lim_{n \to \infty}\sum_{i=1}^{n}2\pi \bar{x}_{i}\delta x \left [ f(\bar{x}_{i}) \right ]=\int_{a}^{b}2 \pi x \delta x [f(x)]dx

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