ASSIGNMENT 4 - SOLUTIONS
1. Compute
ZZ
3x2 y dA,
R
where R is the rectangle [0, 2] × [−1, 0].
Solution. We have
ZZ
Z 0Z
2
3x y dA =
−1
R
2
Z
2
0
h
3x y dx dy =
3
y·x
−1
0
i2
x=0
Z
0
dx =
h i0
8y dy = 4y 2
−1
y=−1
= −4.
2. Let R be the region in the first quadrant bounded by y = 0, y = x and
x2 + y 2 = 1. Compute
ZZ
4
dA.
2
2 2
R (1 + x + y )
Solution. We use polar coordinates
x = r cos θ,
y = r sin θ.
π
Note that the region R corresponds to 0 ≤ θ ≤ , 0 ≤ r ≤ 1. Hence
4
Z πZ 1
Z πZ 2
Z πh
ZZ
i2
4
4
4
4
2
π
4
−1
dθ = ,
dA
=
r
dr
dθ
=
dt
dθ
=
−2t
2
2
2
2
2
2
4
t=1
0 (1 + r )
1 t
0
0
0
R (1 + x + y )
after performing the change of variables t = 1 + r2 (hence 2r dr = dt).
3. Compute
Z 4 Z
0
√
x
−x
y
e
dy dx
x/2
by reversing the order of integration.
Solution. Note that the type 1 region
0 ≤ x ≤√4,
R:
x
x,
2 ≤y ≤
can also be expressed as a region of type 2. Indeed, one can see this by representing R
graphically or reasoning as follows. The second set of inequalities implies x ≤ 2y and
y 2 ≤ x, or y 2 ≤ x ≤ 2y. Also, both sets of inequalities imply 0 ≤ y ≤ 2. Therefore R
is given by
0 ≤ y ≤ 2,
R:
y 2 ≤ x ≤ 2y.
Reversing the order of integration we obtain
Z 2 Z 2y
Z 2h
Z 2
i
x 2y
−x
−
e y dx dy =
−ye y
dy =
(−ye−2 + ye−y ) dy
2
0
y2
x=y
0
1
0
2
ASSIGNMENT 4 - SOLUTIONS
= −e−2
h y 2 i2
2
y=0
Z
+
2
h
i2
y e−y dy = −2e−2 + − ye−y
Z
y=0
0
2
+
e−y dy
0
= −4e−2 − e−2 + 1 = 1 − 5e−2 .
4. Find thepsurface area of the portion of the sphere x2 + y 2 + z 2 = 4 that is inside
the cone z = x2 + y 2 .
Solution. We use spherical coordinates

 x = 2 sin φ cos θ,
y = 2 sin φ sin θ,
r(φ, θ) :

z=
2 cos φ.
Let us now analyze the curve of intersection of the sphere with the cone. Replacing
z 2 = x2 + y 2 in the equation √
of the sphere we obtain z 2 + z 2 = 4. Then z 2 = 2, and
since z ≥ √
0, we obtain z = 2, and
√ hence the intersection curve is a circle in the
plane z = 2. For z = 2 cos φ = 2 we obtain φ = π4 . Then our parameters satisfy
π
0 ≤ φ ≤ and 0 ≤ θ ≤ 2π. For r = r(φ, θ) we get
4
rθ = − 2 sin φ sin θ i + 2 sin φ cos θ j,
rφ = 2 cos φ cos θ i + 2 cos φ sin θ j − 2 sin φ k,
and
i
j
k
rφ × rθ = 2 cos φ cos θ 2 cos φ sin θ −2 sin φ −2 sin φ sin θ 2 sin φ cos θ
0
= 4 sin2 φ cos θ i + 4 sin2 φ sin θ j + 4 sin φ cos φ k.
Therefore
q
krφ × rθ k = 4 sin4 φ cos2 θ + sin4 φ sin2 θ + sin2 φ cos2 φ
q
= 4 sin4 φ + sin2 φ cos2 φ = 4 sin φ
and the area is
ZZ
Z 2π Z π
Z 2π h
iφ= π
4
4
krφ × rθ k dA =
4 sin φ dφ dθ = 4
− cos φ
dθ
R
0
0
0
√ √
2
= 4 · 2π 1 −
= 4 π(2 − 2).
2
φ=0
5. As the site of the world’s largest single-dish radio telescope, the Arecibo Observatory in Puerto Rico is one of the largest centers for research in radio astronomy.
The huge ”dish” is a paraboloidal reflecting surface 305 meters in diameter and 51
meters deep. What is its surface area?
Hint. Find the constant c from the paraboloid’s equation z =
x2 + y 2
.
c
Solution. Let R be the radius of the dish and h its height. We can substitute in
the numerical values 152.5 and 51 at the end. The constant c can be determined by
R2
setting y = R and x = 0 in the paraboloid’s equation. We obtain h =
so that
c
2
R
c=
. To parametrize the paraboloid we use cylindrical coordinates and get
h

x = ρ cos θ


y = ρ sin θ
r(ρ, θ) :

ρ2
 z=
c
ASSIGNMENT 4 - SOLUTIONS
3
with
0 ≤ θ ≤ 2π,
We have
0 ≤ ρ ≤ R.
2
ρ k,
c
rφ = −ρ sin θ i + ρ cos θ j + 0 k,
rθ = cos θ i + sin θ j +
so that
r
krρ × rθ k = ρ
Then the area is
Z
4ρ2
+ 1.
c2
Z 2π Z R2 r
dt
4ρ2
4
ρ
+ 1 dρ dθ =
t + 1 dθ
2
2
c
c
2
0
0
0
0
3 2
3
i
h
i
h
2π
4 2
c2
πc2
4 2
2 c
2
=
R
+
1
−
=
R + 1 − 1 ≈ 80700 square meters.
2
c2
6
6
6
c2
2π
Z
R
r
2
2
Remark Taking into account z = f (x, y) = x +y
one can also compute the area
c
using the formula
ZZ q
1 + fx2 + fy2 dx dy,
Area =
A
where A is the region x2 + y 2 ≤ R2 .