Calculating volume of a solid

Quick Rewind: how to calculate area 1. slice the region to smaller pieces (narrow, nearly rectangle) 2. calculate the area of rectangle 3. take the limit (more and more slices)

Recipe for calculating volume (motivated by area) 1. slice the solid into smaler pieces (thin cut) 2. calculate the volume of each “thin cut” 3. take the limit (more and more slices)

Step 2: volume ≈ area × thickness, because it is so thin

Suppose that x-axis is perpendicular to the direction of cutting, and the solid is between x = a and x = b. Z b Volume= A(x)dx, where A(x) is the area of “thin cut”at x a (“thin cut”is called a cross section, a two-dimensional region)

n Z b X A(x)dx = lim A(yi ) · ∆x, (Riemann sum), where ∆x = (b − a)/n, x0 = a, n→∞ a i=1 x1 = x0 + ∆x, ··· , xk+1 = xk + ∆x, ··· , xn = b, and xi−1 ≤ yi ≤ xi . Slicing method of volume

Example 1: (classical ones) Z h (a) circular cylinder with radius=r, height=h: A(x)dx, A(x) = π[R(x)]2, R(x) = r 0 Z h rx (b) circular cone with radius=r, height=h: A(x)dx, A(x) = π[R(x)]2, R(x) = 0 h Z r √ (c) sphere with radius r: A(x)dx, A(x) = π[R(x)]2, R(x) = r 2 − x2. −r Example 2: Find the volume of the solid obtained by rotating the region bounded by y = x − x2, y = 0, x = 0 and x = 1 about x-axis.

Example 3: Find the volume of the solid obtained by rotating the region bounded by y = x − x2, y = 0, x = 0 and x = 1 about y-axis. Volume of solid of revolution

Disk Method: region between y = f (x), y = 0, a ≤ x ≤ b rotating about x-axis Z b Z b Z d Volume= A(x)dx = π[f (x)]2dx; (about y axis: Volume= π[f (y)]2dy) a a c Washer method: region between y = f (x), y = g(x), a ≤ x ≤ b rotating about x-axis Z b Z b Volume= A(x)dx = π [f (x)]2 − [g(x)]2
dx; a a Z d (about y axis: Volume= π [f (y)]2 − [g(y)]2
dy) c Example 4: Find the volume of the solid obtained by rotating the region bounded by y = x, y = x2 about the y-axis.

Example 5: Find the volume of the solid obtained by rotating the region bounded by y = x, y = x2 about the y = 2.

A more general washer: region between y = f (x), y = g(x), a ≤ x ≤ b rotating Z b Z b about y = k: Volume= A(x)dx = π [f (x) − k]2 − [g(x) − k]2
dx a a Z d (about x = k: Volume= π [f (y) − k]2 − [g(y) − k]2
dy c A(x) = π · [R2(x) − r 2(x)]dx, R(x) = radius of outer circle, r(x) = radius of inner circle Practical Examples

Example 6: An Egypt pharaoh plans to build a pyramid for himself. The base of the pyramid is a square with side 1000 ft, and the height is 300 ft. How much stone does he need to build it? (in ft3)

Example 7: Calculate the volume of a donut. (In mathematics, it is called a torus, see page 372 problem 41)

Summary: Z b 1. Basic formula: A(x)dx, where A(x) is the area of cross section at x a 2. Calculate A(x) from geometric properties (Important cases: A(x) = π[R(x)]2, the disk; and A(x) = π[R2(x) − r 2(x)], the washer.) Calculating volume by (cylindrical) shell method

Example 3: Find the volume of the solid obtained by rotating the region bounded by y = x − x2, y = 0, x = 0 and x = 1 about y-axis.

By washer method: horizontal slice, cross-section: a circular ring with outer circle at √ √ f (y) = (1 + 1 − 4y)/2 and inner circle at g(y) = (1 − 1 − 4y)/2. Z 1/4   V = π [(1 + p1 − 4y)/2]2 − [(1 − p1 − 4y)/2]2 dy 0 By shell method: vertical slice, a thin circular shell is generated by rotating the vertical shell.

For a region bounded by y = f (x) and y = g(x)(f (x) ≥ g(x)), a ≤ x ≤ b, the rotation of a vertical line at x generates a cylinder with radius x and height f (x) − g(x), the surface area of the cylinder is A2(x) = 2πx[f (x) − g(x)]. The volume is Z b Z b V = A2(x)dx = 2πx[f (x) − g(x)]dx a a (cylindrical) shell method

Advantage of shell method (for Example 3): you do not have to solve the inverse functions as in washer method.

For some problems, disk or washer methods are better, while for other situations, shell method is better.

1. Solid by rotating the region between y = f (x) and y = g(x) about x-axis (or y = k):

Z b Washer method: V = π[f 2(x) − g 2(x)]dx; (or a Z b V = π[(f (x) − k)2 − (g(x) − k)2]dx a 2. Solid by rotating the region between y = f (x) and y = g(x) about y-axis (or x = k):

Z b Z b Shell method: V = 2πx[f (x) − g(x)]dx; (or V = 2π(x − k)[f (x) − g(x)]dx a a Example 8 Find the volume of the solid obtained by rotating the region bounded by y = x2 + x − 2, y = 0, about x-axis using (a) shell method (b) slicing method. Which method is better for this problem?