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Surface Area

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Title: Surface Area


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 fx and fy.
We divide D into small rectangles Rij with area
?A ?x ?y. 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 ?Tij of the part of this
tangent plane (a parallelogram) that lies
directly above Rij is an approximation to the
surface area of ?Sij. (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
SURFACE AREA (CONTINUED)
To find the area ?Tij, let a and b denote the
vectors that form the sides of the parallelogram.
Then
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 fx and fy 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 - x2 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 x2 y2 that
lies under the plane z 2.
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