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Surface Area
In this section we will look at the lone application (aside from the area and volume
interpretations) of multiple integrals in this material
...
In other words we were looking at the surface area of
a solid obtained by rotating a function about the x or y axis
...
Here we want to find the surface area of the surface given by z = f ( x, y ) where ( x, y ) is a
point from the region D in the xy-plane
...
Example 1 Find the surface area of the part of the plane 3x + 2 y + z = 6 that lies in the first
octant
...
Let’s first get a sketch of the part of the plane that we are interested in
...
Remember that to get the region D we can pretend that we are standing directly over the plane
and what we see is the region D
...
Plugging z = 0 into the equation of the plane will give us
the equation for the hypotenuse
...
Solution
In this case we are looking for the surface area of the part of z = xy where ( x, y ) comes from
the disk of radius 1 centered at the origin since that is the region that will lie inside the given
cylinder
...
S = ∫∫
x 2 + y 2 + 1 dA
D
=∫
2π
0
∫
1
0
r 1 + r 2 dr dθ
2π
⌠ 1⎛2⎞
=⎮
⎜ ⎟ (1 + r
⌡0 2 ⎝ 3 ⎠
2π
3 1
2 2
⌠ 1 ⎛ 32 ⎞
=⎮
⎜ 2 − 1 ⎟ dθ
⎠
⌡0 3 ⎝
3
⎛
2π 2 ⎞
=
⎜ 2 − 1⎟
3 ⎝
⎠
)
0
dθ