Calculus Volume 1

686

Chapter 6 | Applications of Integration

m = ∑ i =1 n

m i ≈ ∑ i =1 n

ρ ( x i * )Δ x .

This is a Riemann sum. Taking the limit as n →∞, we get an expression for the exact mass of the rod:

→∞ ∑ i =1 n

ρ ( x i * )Δ x = ∫ a b

m = lim n

ρ ( x ) dx .

We state this result in the following theorem.

Theorem 6.7: Mass–Density Formula of a One-Dimensional Object Given a thin rod oriented along the x -axis over the interval ⎡ ⎣ a , b ⎤ the density of the rod at a point x in the interval. Then the mass of the rod is given by

⎦ , let ρ ( x ) denote a linear density function giving

(6.10)

b

m = ∫

ρ ( x ) dx .

a

We apply this theorem in the next example.

Example 6.23 Calculating Mass from Linear Density

Consider a thin rod oriented on the x -axis over the interval [ π /2, π ]. If the density of the rod is given by ρ ( x ) = sin x , what is the mass of the rod?

Solution Applying Equation 6.10 directly, we have m = ∫ a b

π /2 π

ρ ( x ) dx = ∫

π =1.

sin xdx =−cos x | π /2

6.23 Consider a thin rod oriented on the x -axis over the interval [1, 3]. If the density of the rod is given by ρ ( x ) =2 x 2 +3, what is the mass of the rod?

We now extend this concept to find the mass of a two-dimensional disk of radius r . As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object. We assume the density is given in terms of mass per unit area (called area density ), and further assume the density varies only along the disk’s radius (called radial density ). We orient the disk in the xy -plane, with the center at the origin. Then, the density of the disk can be treated as a function of x , denoted ρ ( x ). We assume ρ ( x ) is integrable. Because density is a function of x , we partition the interval from [0, r ] along the x -axis. For i =0, 1, 2,…, n , let P ={ x i } be a regular partition of the interval [0, r ], and for i =1, 2,…, n , choose an arbitrary point x i * ∈ [ x i −1 , x i ]. Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a representative washer are depicted in the following figure.

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