Section 16 6 Surface Area SURFACE AREA Let
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Section 16. 6 Surface Area
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 fx and fy. We divide D into small rectangles Rij with area ΔA = Δx Δy. If (xi, yj) is the corner of Rij closest to the origin, let Pij(xi, yj, f (xi, yj)) be the point on S directly above it. The tangent plane to S and Pij is an approximation to S near Pij. So the area ΔTij of the part of this tangent plane (a parallelogram) that lies directly above Rij is an approximation to the surface area of ΔSij. (See diagrams on page 1056 of the text. )
SURFACE AREA (CONTINUED) The surface area of S is denoted by A(S) and is given by
SURFACE AREA (CONTINUED) To find the area ΔTij, let a and b denote the vectors that form the sides of the parallelogram. Then The area of the parallelogram is |a × b|
SURFACE AREA (CONTINUED) So,
SURFACE AREA (CONCLUDED) The total surface area, A(S), of the surface S with equation z = f (x, y), , where fx and fy are continuous, is given by
SURFACE AREA IN LEIBNIZ NOTATION
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.
