1. Section 16.6 Surface Area

2. SURFACE AREA Let S be a surface defined by z = f (x, y) over a specified region D. Assume that f has continuous first partial derivatives f x and f y. We divide D into small rectangles R ij with area ΔA = Δx Δy. 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. The tangent plane to S and P ij is an approximation to S near P ij. So the area ΔT ij of the part of this tangent plane (a parallelogram) that lies directly above R ij is an approximation to the surface area of ΔS ij. (See diagrams on page 1056 of the text.)

3. SURFACE AREA (CONTINUED) The surface area of S is denoted by A(S) and is given by

4. To find the area ΔT ij, let a and b denote the vectors that form the sides of the parallelogram. Then SURFACE AREA (CONTINUED) The area of the parallelogram is |a × b|

5. SURFACE AREA (CONTINUED) So,

6. SURFACE AREA (CONCLUDED) The total surface area, A(S), of the surface S with equation z = f (x, y),, where f x and f y are continuous, is given by

7. SURFACE AREA IN LEIBNIZ NOTATION

8. EXAMPLES 1.Find the surface area of the part of the surface z = 1 − x 2 + y that lies above the triangular region D in the xy-plane with vertices (1, 0), (0, −1) and (0, 1) 2.Find the area of the part of the paraboloid z = 1 + x 2 + y 2 that lies under the plane z = 2.