Volume Consider a solid for which one wants the volume. Suppose an x - - PDF document

Volume

Consider a solid for which one wants the volume. Suppose an x− axis is drawn and it is found that for a particular value of x the cross- sectional area is given by a function A(x). This means that if one slices the solid by making a cut through the solid perpendicular to the x− axis, the cut goes through a portion of the solid along a plane (the cross-section) and the area of that portion is A(x).

Volume of a Slice

Now, suppose one cuts the solid into thin slices by cutting perpen- dicular to the points { x0, x1, x2, . . . , xn} along the x− axis. This deter- mines a partition P of [a, b], where a = x0 and b = xn. The volume ∆Vk of the kth slice may be approximated by taking the area A(ck) of the cross-section anywhere in the interval [xk−1, xk] and multiplying by the thickness ∆xk of the slice. So ∆Vk ≈ A(ck)∆xk. The total volume is thus n

k=1 ∆Vk ≈ n k=1 A(ck)∆xk.

This is the Riemann Sum R(A, P, a, b) and we may conclude the vol- ume is equal to the integral b

a A(x) dx.

Volume of a Solid of Revolution

Suppose a solid is obtained by taking a region of the form { (x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)} and rotating it about the x− axis. The solid

• btained is called a Solid of Revolution.

Each cross section is a circle of radius f(x), so the cross-sectional area is given by πf(x)2 and the volume will be b

a πf(x)2 dx = π

b

a f(x)2 dx.

This is sometimes referred to as the Disk Method, since the individual slices look like disks.

Variations

• One may also take a region bounded by a curve of the form

x = g(y), two horizontal lines and the y− axis and rotate it about the y− axis. One gets a similar formula, with the roles of x and y interchanged.

• One may rotate around a line parallel to one of the coordinate
• axes. In this case, the key point to remember is the cross-section

is a circle so one must figure out what the radius is.

Cylindrical Shells

Suppose one creates a solid by taking a region of the form { (x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)} but rotating it about the y− axis rather

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than around the x− axis. One may still be able to find its volume by employing the Disk Method in ingenious ways, but it will usually be easier to use another method called the Method of Cylindrical Shells. To make the analysis simpler, assume a ≥ 0, so the region is in the first quadrant, and create a partition P = { x0, x1, x2, . . . , xn} of [a, b]. Let Sk be the solid obtained by rotating the vertical strip { (x, y) : xk−1 ≤ x ≤ xk} about the y− axis and let ∆Vk be the volume of Sk. The volume of the entire solid will be the sum n

k=1 ∆Vk. We will

estimate the entire volume by finding an estimate for each ∆Vk and adding them up. This will give us a Riemann Sum.

Volume of an Individual Shell

Sk may be visualized as almost being a cylindrical shell. It’s rela- tively simple to visualize the volume of a cylindrical shell of a given height, radius and thickness as follows. Visualize making a vertical cut through the height of the shell. Many tin cans have seams; one may imagine cutting along the seam. Now separate the edges along the cut and flatten the shell. You’ll

• btain a very thin rectangular solid whose volume, which is obviously

equal to the volume of the original shell, will be equal to the product

• f its length, width and height.

The height of the rectangular solid is clearly the same as the height of the shell and the width is clearly the same as the thickness of the shell. The length of the rectangular solid is clearly equal to the circumference

• f the shell, which is 2π× the radius of the shell.

Thus the volume of the solid, and hence the shell, will equal the product

• f the height, thickness and 2π× its radius.

Sk is not quite a cylindrical shell, but this calculation may be used to estimate its volume. The height of Sk varies, but if we choose some ck in the interval [xk−1, xk] the height will not vary much from f(ck). The thickness of Sk is clearly ∆xk. The radius of Sk is not unambiguously defined, but it certainly isn’t much different from ck. We thus get ∆Vk ≈ f(ck) · ∆xk · 2πck = 2πckf(ck)∆xk. We may thus approximate the volume by n

k=1 2πckf(ck)∆xk

= 2π n

k=1 ckf(ck)∆xk.

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This is 2πR(xf(x), P, a, b) and we conclude the volume is the integral 2π b

a xf(x) dx.

Variations

As with the Disk Method, there will be variations. The key point to remember is to visualize a cylindrical shell and pick out the radius, height and thickness. It is the thickness which determines whether one integrates with respect to x or with respect to y. If one obtains the shells by partitioning an interval along the y− axis, then one will need to integrate with respect to y.