15.5 Surface Area

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15.5 Surface Area

Surface Area

In this section we apply double integrals to the problem of computing the area of a surface.

Here we compute the area of a surface with equation z = f(x, y), the graph of a function of two variables.

Let S be a surface with equation z = f(x, y), where f has continuous partial derivatives.

For simplicity in deriving the surface area formula, we

assume that f(x, y) ≥ 0 and the domain D of f is a rectangle.

3

Surface Area

If (x_i, y_j) is the corner of R_ij closest to the origin, let P_ij (x_i, y_j, f(x_i, y_j)) be the point on S directly above it (see Figure 1).

Figure 1

Surface Area

The tangent plane to S at P_ij is an approximation to S near P_ij. So the area ∆T_ij of the part of this tangent plane

(a parallelogram) that lies directly above R_ij is an

approximation to the area ∆S_ij of the part of S that lies directly above R_ij.

Thus the sum ΣΣ∆T_ij is an approximation to the total area of S, and this approximation appears to improve as the number of rectangles increases.

Therefore we define the surface area of S to be

5

Surface Area

To find a formula that is more convenient than Equation 1 for computational purposes, we let a and b be the vectors that start at P_ij and lie along the sides of the parallelogram with area ∆T_ij. (See Figure 2.)

Figure 2

Surface Area

Then ∆T_ij = |a × b|. We know that f_x(x_i, y_j) and f_y(x_i, y_j) are the slopes of the tangent lines through P_ij in the directions of a and b.

Therefore

a = ∆xi + f_x(x_i, y_j) ∆xk b = ∆yj + f_y(x_i, y_j) ∆yk and

7

Surface Area

= –f_x(x_i, y_j) ∆x ∆yi – f_y(x_i, y_j) ∆x ∆yj + ∆x ∆yk

= [–f_x(x_i, y_j)i – f_y(x_i, y_j)j + k] ∆A Thus

∆T_ij = |a × b| = ∆A From Definition 1 we then have

Surface Area

and by the definition of a double integral we get the following formula.

9

Surface Area

If we use the alternative notation for partial derivatives, we can rewrite Formula 2 as follows:

Notice the similarity between the surface area formula in Equation 3 and the arc length formula

Example 1

Find the surface area of the part of the surface z = x² + 2y that lies above the triangular region T in the xy-plane with vertices (0, 0), (1, 0), and (1, 1).

Solution:

The region T is shown in Figure 3 and is described by T = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x}

11

Example 1 – Solution

Using Formula 2 with f(x, y) = x² + 2y, we get

cont’d

Example 1 – Solution

Figure 4 shows the portion of the surface whose area we have just computed.

cont’d