Math 1C Section 7220 Spring 2017
Instructor: John Kwon
Math 1C Special Topic: Surface Areas - Solution
1. Find the area of the paraboloid z = 25 − x2 − y 2 that lies above the xy-plane.
Solution. Let S be the portion of the paraboloid z = 25 − x2 − y 2 that lies above the xy-plane.
We are to compute
ZZ q
Area(S) =
(fx )2 + (fy )2 + 1 dA
R
25 − x2
where f (x, y) =
fy = −2y, we have
− y2
ZZ p
and R is the projection of S onto the xy-plane. Since fx = −2x and
(fx
)2
+ (fx
)2
ZZ p
+ 1 dA =
R
4x2 + 4y 2 + 1 dA .
R
Since the intersection
of the paraboloid
and the xy-plane is 25 − x2 − y 2 = 0, or x2 + y 2 = 25,
we see that R = (x, y) | x2 + y 2 ≤ 25 . Therefore we use polar coordinate to set up the double
integral as:
Z 2π Z 5 p
ZZ p
2
2
4x + 4y + 1 dA =
4r2 + 1 · r dr dθ.
0
R
0
This integral can be done by u-substitution with
u = 4r2 + 1,
so that we have
Z 2π Z 5 p
0
0
4r2
Z
Z
1 2π 101 1/2
u du dθ
+ 1 · 8r dr dθ =
8 0
0
0
1
Z
Z 2π 1
1 2π 2 3/2 u=101
dθ =
=
u 1013/2 − 1 dθ
8 0 3
12 0
u=1
√
√
π 101 101 − 1
1
.
=
· 2π 101 101 − 1 =
12
6
1
+ 1 · r dr dθ =
8
Z
2π
du = 8r dr
Z
5p
4r2
2. Find the area of the surface that is the part of the surface z = 5 + 2x2 + 3y that lies above
the triangle on the xy-plane with vertices (0, 0), (1, 0), and (1, 2).
Solution. Let S be the part of the surface z = 5 + 2x2 + 3y that lies above the triangle on the
xy-plane with vertices (0, 0), (1, 0), and (1, 2). We are to compute
ZZ q
Area(S) =
(fx )2 + (fy )2 + 1 dA
R
where R is the triangle on the xy-plane with vertices (0, 0), (1, 0), and (1, 2). By drawing a
picture, we see that R is bounded by the x-axis and the graphs of x = 1 and y = 2x. Since
f (x, y) = 5 + 2x2 + 3y, we have fx = 4x and fy = 3, and hence
(Continued on the next page)
1 Z 2x p
ZZ p
Z
ZZ q
2
2
2
(fx ) + (fy ) + 1 dA =
16x + 10 dA =
R
R
0
We now compute the iterated integral as follows:
Z 1 Z 2x p
Z 1p
Z
2
2
16x + 10 dy dx =
16x + 10
0
0
16x2 + 10 dy dx .
0
0
2x
Z
dy dx =
0
1p
16x2 + 10 · 2x dx
0
where the u-substitution
u = 16x2 + 10,
may be used so that we have
Z 1 Z 2x p
Z
2
16x + 10 dy dx =
0
0
1p
16x2
du = 32x dx
Z
26
u1/2 du
+ 10 · 2x dx =
0
10
26
√ √
2 26 26 − 10 10
2u3/2 =
=
=
3 3
10
√ √
4 13 26 − 5 10
.
3
3. Find the area of the surface that is the part of the plane 4x + 3y + z = 12 that lies inside
the cylinder x2 + y 2 = 4.
Solution. Let S be the part of the plane 4x + 3y + z = 12 (Note: solving this for z yields
z = 12 − 4x − 3y) that lies inside the cylinder x2 + y 2 = 4. We are to compute
ZZ q
Area(S) =
(fx )2 + (fy )2 + 1 dA
R
where f (x, y) = 12 − 4x − 3y. Therefore fx = −4 and fy = −3, so we have
ZZ q
ZZ √
2
2
Area(S) =
(fx ) + (fy ) + 1 dA =
26 dA.
R
R
Since R = (x, y) | x2 + y 2 ≤ 4 , which is a disk on the xy-plane centered at the origin and radius
2, we could use the polar coordinates to set up and calculate the double integral:
ZZ √
R
r=2
r2 26 dA =
26 · r dr dθ = 26
dθ
2 r=0
0
0
0
Z 2π
√
√
√
= 26 · 2
dθ = 26 · 2 · 2π = 4π 26 .
Z
2π
Z
2√
√
Z
2π
0
Alternatively, we could have simply used the fact that
ZZ
dA = Area(R) = π · 22 = 4π
R
and computed
Area(S) =
ZZ √
R
26 dA =
√
ZZ
26
dA =
R
√
√
26 · 4π = 4π 26 .
4. Find the area of the surface that is the part of the sphere x2 + y 2 + z 2 = 16 that lies
within the cylinder x2 + y 2 = 4x.
Solution. Let S be the part of the sphere x2 + y 2 + z 2 = 16 that lies within the cylinder
x2 + y 2 = 4x. As in the previous problems, we are to compute the area of S using a double
integral, but this time the setting is a bit different. Note that the cylinder x2 + y 2 = 4x intersects
the sphere x2 + y 2 + z 2 = 16 in two different pieces (one above the xy-plane, and one below), so
we should set up the usual integral to find the surface area of the piece above the xy-plane and
double the result; that is,
ZZ q
Area(S) = 2 ·
(fx )2 + (fy )2 + 1 dA
R
where R is the region on the xy-plane inside the curve x2 + y 2 = 4x (which is a p
circle of radius
2 centered at (2, 0)). Also, solving the equation of
the
sphere
for
z
yields
z
=
±
16 − x2 − y 2 ,
p
where we take the positive part so that f (x, y) = 16 − x2 − y 2 . Then we have
fx = p
fy = p
−x
,
16 − x2 − y 2
−y
,
16 − x2 − y 2
so we have
s
y2
x2
+
+1 =
16 − x2 − y 2 16 − x2 − y 2
r
16
4
= p
16 − x2 − y 2
16 − x2 − y 2
q
(fx )2 + (fy )2 + 1 =
=
s
x2 + y 2 + 16 − x2 − y 2
16 − x2 − y 2
so the surface area we must compute becomes
ZZ q
ZZ
4
2
2
p
Area(S) = 2 ·
dA .
(fx ) + (fy ) + 1 dA = 2 ·
16 − x2 − y 2
R
R
As for R, note that the equation x2 + y 2 = 4x can be written in the following polar form:
x2 + y 2 = 4x
=⇒ r2 = 4r cos θ
=⇒ r = 4 cos θ
so the double integral over R can be set up using the polar coordinates:
ZZ
Area(S) = 2 ·
R
4
Z
p
dA = 2 ·
16 − x2 − y 2
0
π
Z
0
4 cos θ
√
4
r dr dθ
16 − r2
where the upper limit of integration for θ is only up to π because one loop of the circle r = 4 cos θ
is completed from θ = 0 to θ = π. Now we use the u-substitution
u = 16 − r2 ,
du = −2r dr
to compute the above integral as follows:
(Continued on the next page)
Z
π
Z
4 cos θ
Z
0
π
Z
Z π Z 4 cos θ
4
4
√
r dr dθ = −
2r dr dθ
16 − r2
16 − r2
0
0
Z π Z 16
4
8
√ du dθ =
√ du dθ
2
u
2
u
Z π
Z 0π 16 sin θ
(1 − sin θ) dθ
(4 − 4 sin θ) dθ = 32
dθ = 8
2
√
Area(S) = 2 ·
0
16 sin2 θ
= −
Z π 0 16 √ u=16
8 u
=
u=16 sin θ
0
0
0
π
= 32(θ + cos θ) = 32 [(π + cos π) − (0 + cos 0)] = 32(π − 2) .
0
5. Prove that the area of the part√of the plane z = ax + by + c that projects onto a region D
in the xy-plane with area A(D) is a2 + b2 + 1 A(D).
Proof. Let S be the part of the plane z = ax + by + c that projects onto D in the xy-plane. Then
as usual
ZZ q
Area(S) =
(fx )2 + (fy )2 + 1 dA, ,
D
where f (x, y) = ax + by + c, so fx = a and fy = b (here a, b, and c are constants). Therefore
ZZ p
ZZ q
2
2
(fx ) + (fy ) + 1 dA =
a2 + b2 + 1 dA
Area(S) =
D
D
but since
√
a2 + b2 + 1 is a constant, we have
ZZ p
ZZ
p
2
2
2
2
a + b + 1 dA =
a +b +1
1 dA .
D
D
ZZ
Finally,
1dA = A(D), so we have
D
ZZ
p
p
2
2
a +b +1
1 dA =
a2 + b2 + 1 A(D)
Area(S) =
D
as desired.