Multivariate Calculus - Bruce E. Shapiro

LECTURE 19. SURFACE AREA WITH DOUBLE INTEGRALS 157
Example 19.1 Find the surface area of the region of the plane 3x − 2y + 6z = 12
that is bounded by the planes x = 0, y = 0 and 3x + 2y = 12.
Solution The domain is the region in the xy plane bounded by the x-axis, the y-axis,
and the line
3x + 2y = 12
Solving the line for y,
y = 6 − 1.5x
This line intersects the x-axis at x = 12/3 = 4, and the y-axis at y = 12/2 = 6.
Therefore we can write the domain of the integral as
And the surface area as
A =
{(x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 6 − 1.5x}
∫ 4 ∫ 6−1.5x
0
0
√
1 + f 2 x(x, y) + f 2 y (x, y)dydx
where f(x, y) is the solution of 3x − 2y + 6z = 12 for z = f(x, y), namely
f(x, y) = 2 − x 2 + y 3
Differentiating, f x = −1/2 and f y = 1/3, thus
A =
=
∫ 4 ∫ 6−1.5x
0 0
∫ 4 ∫ 6−1.5x
0
0
= √ 49/36
= 7 6
= 7 6
√
1 + f 2 x(x, y) + f 2 y (x, y)dydx
√
1 + (−1/2) 2 + (1/3) 2 dydx
∫ 4 ∫ 6−1.5x
0 0
∫ 4 ∫ 6−1.5x
0
∫ 4
0
0
dydx
(6 − 1.5x)dx
= 7 6 (6x − 0.75x2 )| 4 0
= 7 (24 − 12) = 14
6
dydx
Example 19.2 Find the surface area of the part of the surface
that lies above the quarter of the circle
in the first quadrant.
z = √ 9 − y 2
x 2 + y 2 = 9
Math 250, Fall 2006 Revised December 6, 2006.