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Surface Integrals
It is now time to think about integrating functions over some surface, S, in three-dimensional
space
...
Here is a sketch of some surface S
...
We used a
rectangle here, but it doesn’t have to be of course
...
Do not get so
locked into the xy-plane that you can’t do problems that have regions in the other two planes
...
There are essentially two separate methods here, although as we will see they are really the same
...
In this
case the surface integral is,
∫∫
S
2
2
⌠⌠
⎛ ∂g ⎞ ⎛ ∂g ⎞
f ( x, y, z ) dS = ⎮⎮ f ( x, y, g ( x, y ) ) ⎜ ⎟ + ⎜ ⎟ + 1 dA
⎮⎮
⎝ ∂x ⎠ ⎝ ∂y ⎠
⌡⌡
D
Now, we need to be careful here as both of these look like standard double integrals
...
The integral on the left however is a surface
integral
...
The surface integral will
have a dS while the standard double integral will have a dA
...
After that the integral is a standard double
integral and by this point we should be able to deal with that
...
We will see one of these formulas in the
examples and we’ll leave the other to you to write down
...
Before we work some examples let’s notice that since we can parameterize a surface given by
z = g ( x, y ) as,
r ( x, y ) = xi + yj + g ( x, y ) k
we can always use this form for these kinds of surfaces as well
...
You might want to verify this for the practice of computing these
cross products
...
Example 1 Evaluate
∫∫ 6 xy dS
where S is the portion of the plane x + y + z = 1 that lies in
S
front of the yz-plane
...
This is easy enough
to do
...
Here is a sketch of the surface S
...
Notice that the axes are labeled differently than we are used to seeing in the sketch of D
...
We arrived at the equation of the
hypotenuse by setting x equal to zero in the equation of the plane and solving for z
...
0 ≤ y ≤1
0 ≤ z ≤ 1− y
Now, because the surface is not in the form z = g ( x, y ) we can’t use the formula above
...
Here it is,
∫∫
S
2
2
⌠⌠
⎛ ∂g ⎞ ⎛ ∂g ⎞
f ( x, y, z ) dS = ⎮⎮ f ( g ( y, z ) , y, z ) 1 + ⎜ ⎟ + ⎜ ⎟ dA
⎮⎮
∂
∂
y
z
⎝
⎠
⎝
⎠
⌡⌡
D
The changes made to the formula should be the somewhat obvious changes
...
∫∫ 6 xy dS = ∫∫ 6 (1 − y − z ) y
S
1 + ( −1) + ( −1) dA
2
2
D
Notice that we plugged in the equation of the plane for the x in the integrand
...
Here is that work
...
S
Solution
We gave the parameterization of a sphere in the previous section
...
r (θ , ϕ ) = 2sin ϕ cos θ i + 2sin ϕ sin θ j + 2 cos ϕ k
Since we are working on the upper half of the sphere here are the limits on the parameters
...
Here are the two individual vectors
...
i
rθ × rϕ = −2sin ϕ sin θ
2 cos ϕ cos θ
j
2sin ϕ cos θ
2 cos ϕ sin θ
k
0
−2sin ϕ
= −4sin 2 ϕ cos θ i − 4sin ϕ cos ϕ sin 2 θ k − 4sin 2 ϕ sin θ j − 4sin ϕ cos ϕ cos 2 θ k
= −4sin 2 ϕ cos θ i − 4sin 2 ϕ sin θ j − 4sin ϕ cos ϕ ( sin 2 θ + cos 2 θ ) k
= −4sin 2 ϕ cos θ i − 4sin 2 ϕ sin θ j − 4sin ϕ cos ϕ k
Finally, we need the magnitude of this,
rθ ×rϕ = 16sin 4 ϕ cos 2 θ + 16sin 4 ϕ sin 2 θ + 16sin 2 ϕ cos 2 ϕ
= 16sin ϕ ( cos θ + sin θ ) + 16sin ϕ cos ϕ
4
2
2
2
2
= 16sin 2 ϕ ( sin 2 ϕ + cos 2 ϕ )
= 4 sin ϕ
2
= 4 sin ϕ
= 4sin ϕ
We can drop the absolute value bars in the sine because sine is positive in the range of ϕ that we
are working with
...
Here is the evaluation for the double integral
...
Solution
We parameterized up a cylinder in the previous section
...
r ( z ,θ ) = 3 cos θ i + 3 sin θ j + z k
The ranges of the parameters are,
0≤ z≤6
0 ≤ θ ≤ 2π
Now we need rz × rθ
...
rz ( z ,θ ) = k
rθ ( z ,θ ) = − 3 sin θ i + 3 cos θ j
Here is the cross product
...
Solution
There is a lot of information that we need to keep track of here
...
However, unlike
the previous example we are putting a top and bottom on the surface this time
...
Actually we need to be careful here
...
We’re going to let S1 be the portion of the cylinder that goes from the xy-plane to the plane
...
We’ll call the portion of the plane that
lies inside (i
...
the cap on the cylinder) S 2
...
In order to do this integral we’ll need to note that just like the standard double integral, if the
surface is split up into pieces we can also split up the surface integral
...
However, we’ve done most of the work for the
first one in the previous example so let’s start with that
...
Here
they are
...
However, since we are on the
cylinder we know what y is from the parameterization so we will also need to plug that in
...
∫∫ y + z dS = ∫∫ (
S1
3 sin θ + z
)( 3 ) dA
D
= 3∫
2π
0
∫
4 − 3 sin θ
3 sin θ + z dz dθ
0
(
2π
) (
1
= 3∫
3 sin θ 4 − 3 sin θ + 4 − 3 sin θ
0
2
2π
3 2
= 3 ∫ 8 − sin θ dθ
0
2
2π
3
= 3 ∫ 8 − (1 − cos ( 2θ ) ) dθ
0
4
)
2
dθ
2π
3
⎛ 29
⎞
= 3 ⎜ θ + sin ( 2θ ) ⎟
4
⎝ 4
⎠0
29 3 π
=
2
S2 : Plane on Top of the Cylinder
In this case we don’t need to do any parameterization since it is set up to use the formula that we
gave at the start of this section
...
Also note that,
3 centered at the origin
...
∫∫ y + z dS = ∫∫ ( y + 4 − y ) ( 0 ) + ( −1)
2
S2
2
+ 1 dA
D
= 2 ∫∫ 4 dA
D
Don’t forget that we need to plug in for z! Now at this point we can proceed in one of two ways
...
Here is the remainder of the work for this problem
...
Also, don’t forget to plug in for z
...
∫∫ y + z dS = ∫∫ ( y + 0 ) ( 0 ) + ( 0 ) + (1)
2
S3
2
2
dA
D
= ∫∫ y dA
D
2π
=⌠
⌡0
∫
0
3
r 2 sin θ dr dθ
2π
3
⌠ ⎛1 3
⎞
= ⎮ ⎜ r sin θ ⎟ dθ
⎠0
⌡0 ⎝ 3
=∫
2π
0
3 sin θ dθ
= − 3 cos θ
2π
0
0
We can now get the value of the integral that we are after