Test 3 Review
Problem 1 (c) (1999):
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
y2 ≤ x ≤ 1
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
y2 ≤ x ≤ 1
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
y2 ≤ x ≤ 1
Z
Z
cos(x3/2) dy dx = ?
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
y2 ≤ x ≤ 1
Z 1Z
0
cos(x3/2) dy dx = ?
Problem 1 (c) (1999):
Z 1 Z 1
−1 y 2
cos(x3/2) dx dy = ?
−1 ≤ y ≤ 1
y2 ≤ x ≤ 1
Z 1 Z √x
0
3/2) dy dx = ?
cos(x
√
− x
Z 1 Z √x
0
3/2) dy dx
cos(x
√
− x
Z 1 Z √x
0
3/2) dy dx
cos(x
√
− x
Z 1
√
=
2 x cos(x3/2) dx
0
Z 1 Z √x
0
3/2) dy dx
cos(x
√
− x
Z 1
√
=
2 x cos(x3/2) dx
0
1
4
3/2
= sin(x ) 3
0
Z 1 Z √x
0
3/2) dy dx
cos(x
√
− x
Z 1
√
=
2 x cos(x3/2) dx
0
1
4
3/2
= sin(x ) 3
0
=
4
3
sin(1)
Z 1 Z √x
0
3/2) dy dx
cos(x
√
− x
Z 1
√
=
2 x cos(x3/2) dx
0
1
4
3/2
= sin(x ) 3
0
=
4
3
sin(1)
Problem 3 (1999):
Problem 3 (1999):
(
3
f (x, y) = −1
2
if 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2
if 0 ≤ x ≤ 2 and 2 < y ≤ 3
if 2 < x ≤ 3 and 0 ≤ y ≤ 3.
Problem 3 (1999):
(
3
f (x, y) = −1
2
if 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2
if 0 ≤ x ≤ 2 and 2 < y ≤ 3
if 2 < x ≤ 3 and 0 ≤ y ≤ 3.
ZZ
f (x, y) dA =
R
ZZ
f (x, y) dA = 4 × 3 + 2 × (−1) + 3 × 2 = 16
R
Z 3Z 1
f (x, y) dy dx =
0
0
Z 3Z 1
f (x, y) dy dx =
0
0
Z 3Z 1
f (x, y) dy dx = 2 × 3 + 1 × 2 = 8
0
0
Problem 4 (1999):
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
Calculate (r, θ, z) and (ρ, θ, φ).
√
2, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
Calculate (r, θ, z) and (ρ, θ, φ).
r2 =
√
2, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4
√
2, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4
so
r=2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4
x=y>0
so
r=2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4
x = y > 0 so
so r = 2
θ = π/4
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4 so r = 2
x = y > 0 so θ = π/4
z = z so z = 2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
r 2 = x2 + y 2 = 4 so r = 2
x = y > 0 so θ = π/4
z = z so z = 2
(r, θ, z) = (2, π/4, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 =
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
so
√
ρ=2 2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
θ=
so
√
ρ=2 2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
θ=θ
so
√
so ρ = 2 2
θ = π/4
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
θ=θ
z=
so
√
so ρ = 2 2
θ = π/4
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
θ=θ
z = ρ cos φ,
so
√
so ρ = 2 2
θ = π/4
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
√
so ρ = 2 2
θ = θ so θ = π/4
√
z = ρ cos φ, cos φ = 1/ 2
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
√
so ρ = 2 2
θ = θ so θ = π/4
√
z = ρ cos φ, cos φ = 1/ 2 so φ = π/4
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
ρ2 = x2 + y 2 + z 2 = 8
√
so ρ = 2 2
θ = θ so θ = π/4
√
z = ρ cos φ, cos φ = 1/ 2 so φ = π/4
√
(ρ, θ, φ) = (2 2, π/4, π/4)
Problem 4 (1999):
√
(x, y, z) = ( 2,
√
2, 2)
Calculate (r, θ, z) and (ρ, θ, φ).
(r, θ, z) = (2, π/4, 2)
√
(ρ, θ, φ) = (2 2, π/4, π/4)
Problem 6 (1999):
Problem 6 (1999):
Express the volume of the solid above the surface z =
x2 + y 2 and below the surface x2 + y 2 + z 2 = 6 as an
iterated integral in polar or cylindrical coordinates. Do not
evaluate the integral.
Problem 6 (1999):
Express the volume of the solid above the surface z =
x2 + y 2 and below the surface x2 + y 2 + z 2 = 6 as an
iterated integral in polar or cylindrical coordinates. Do not
evaluate the integral.
Problem 6 (1999):
Express the volume of the solid above the surface z =
x2 + y 2 and below the surface x2 + y 2 + z 2 = 6 as an
iterated integral in polar or cylindrical coordinates. Do not
evaluate the integral.
z = x2 + y 2
x2 + y 2 + z 2 = 6
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
(z + 3)(z − 2) = 0
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
(z + 3)(z − 2) = 0
z=2
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
(z + 3)(z − 2) = 0
z=2
The surfaces intersect on the plane z = 2.
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
(z + 3)(z − 2) = 0
z=2
The surfaces intersect on the plane z = 2.
The projection of this intersection to the xy-plane is
z = x2 + y 2
x2 + y 2 + z 2 = 6
z + z2 = 6
z2 + z − 6 = 0
(z + 3)(z − 2) = 0
z=2
The surfaces intersect on the plane z = 2.
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z
Z
Z
dz dr dθ
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z
Z
Z
r dz dr dθ
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z 2π Z
Z
r dz dr dθ
0
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z 2π Z √2 Z
r dz dr dθ
0
0
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z
Z √ Z √
0
6−r 2
2
2π
0
r2
r dz dr dθ
z = x2 + y 2
x2 + y 2 + z 2 = 6
The projection of this intersection to the xy-plane is
the circle x2 + y 2 = 2.
Express the volume using polar or cylindrical coordinates.
Z 2π Z √2 Z √6−r2
0
0
r2
r dz dr dθ
Problem 7 (1999):
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z
Z
Z
dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z
Z
Z
ρ2 sin φ dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z
Z
Z
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z π/2 Z
0
Z
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z π/2 Z 2π Z
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z π/2 Z 2π Z 1
0
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Problem 7 (1999):
Using spherical coordinates, calculate
ZZZ
3/2
2
2
2
x +y +z
dV
S
where S is the solid bounded above by the sphere x2 +
y 2 + z 2 = 1 and below by the plane z = 0.
Z π/2 Z 2π Z 1
0
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Z π/2 Z 2π Z 1
0
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
0
0
0
ρ5 sin φ dρ dθ dφ
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
0
=
0
0
Z π/2 Z 2π
1
ρ5 sin φ dρ dθ dφ
6 0
sin φ dθ dφ
0
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
ρ5 sin φ dρ dθ dφ
0
=
0
0
Z π/2 Z 2π
1
sin φ dθ dφ
6 0
0
Z π/2
π
=
sin φ dφ
3 0
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
ρ5 sin φ dρ dθ dφ
0
=
0
0
Z π/2 Z 2π
1
sin φ dθ dφ
6 0
0
Z π/2
π
=
sin φ dφ
3 0
π/2
π
= (− cos φ)
3
0
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
ρ5 sin φ dρ dθ dφ
0
=
0
0
Z π/2 Z 2π
1
sin φ dθ dφ
6 0
0
Z π/2
π
=
sin φ dφ
3 0
π/2
π
π
=
= (− cos φ)
3
3
0
Z π/2 Z 2π Z 1
0
0
(ρ2)3/2 ρ2 sin φ dρ dθ dφ
0
Z π/2 Z 2π Z 1
=
ρ5 sin φ dρ dθ dφ
0
=
0
0
Z π/2 Z 2π
1
sin φ dθ dφ
6 0
0
Z π/2
π
=
sin φ dφ
3 0
π/2
π
π
=
= (− cos φ)
3
3
0
Other Answers From 1999:
Other Answers From 1999:
Problem (1)(a): 1/2
Other Answers From 1999:
Problem (1)(a): 1/2
Problem (1)(b): π 2
Other Answers From 1999:
Problem (1)(a): 1/2
Problem (1)(b): π 2
Z 1Z 1
Problem (2)(a):
f (x, y) dx dy
0
y
Other Answers From 1999:
Problem (1)(a): 1/2
Problem (1)(b): π 2
Z 1Z 1
Problem (2)(a):
f (x, y) dx dy
0
y
Z 1 Z y 1/3
f (x, y) dx dy
Problem (2)(b):
0
0
Other Answers From 1999:
Problem (1)(a): 1/2
Problem (1)(b): π 2
Z 1Z 1
Problem (2)(a):
f (x, y) dx dy
0
y
Z 1 Z y 1/3
f (x, y) dx dy
Problem (2)(b):
0
0
Z 3 Z 2 Z √4−y 2
dz dy dx
Problem (5):
0
0
0
Quick Overview:
Quick Overview:
• For polar and cylindrical coordinates,
x2 + y 2 = r 2 ,
x = r cos θ,
y = r sin θ.
Quick Overview:
• For polar and cylindrical coordinates,
x2 + y 2 = r 2 ,
x = r cos θ,
y = r sin θ.
• For spherical coordinates,
x2 +y 2 +z 2 = ρ2, z = ρ cos φ, r = ρ sin φ.
Quick Overview:
• For polar and cylindrical coordinates,
x2 + y 2 = r 2 ,
x = r cos θ,
y = r sin θ.
• For spherical coordinates,
x2 +y 2 +z 2 = ρ2, z = ρ cos φ, r = ρ sin φ.
• The value of r is the distance to the z-axis and the value
of ρ is the distance to the origin.
Quick Overview:
• For polar and cylindrical coordinates,
x2 + y 2 = r 2 ,
x = r cos θ,
y = r sin θ.
• For spherical coordinates,
x2 +y 2 +z 2 = ρ2, z = ρ cos φ, r = ρ sin φ.
• The value of r is the distance to the z-axis and the value
of ρ is the distance to the origin.
• For polar coordinates, remember dA = r dr dθ.
Quick Overview:
• For polar and cylindrical coordinates,
x2 + y 2 = r 2 ,
x = r cos θ,
y = r sin θ.
• For spherical coordinates,
x2 +y 2 +z 2 = ρ2, z = ρ cos φ, r = ρ sin φ.
• The value of r is the distance to the z-axis and the value
of ρ is the distance to the origin.
• For polar coordinates, remember dA = r dr dθ.
• For cylindrical coordinates, remember
dV = r dz dr dθ.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• When doing triple integrals involving the expression
x2 + y 2, I will think “cylindrical coordinates”.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• When doing triple integrals involving the expression
x2 + y 2, I will think “cylindrical coordinates”.
• When doing triple integrals involving the expression
x2 + y 2 + z 2, I will think “spherical coordinates”.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• When doing triple integrals involving the expression
x2 + y 2, I will think “cylindrical coordinates”.
• When doing triple integrals involving the expression
x2 + y 2 + z 2, I will think “spherical coordinates”.
• I will practice interchanging the order of integration.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• When doing triple integrals involving the expression
x2 + y 2, I will think “cylindrical coordinates”.
• When doing triple integrals involving the expression
x2 + y 2 + z 2, I will think “spherical coordinates”.
• I will practice interchanging the order of integration.
• I will practice setting up integrals for the test.
• For spherical coordinates, remember
dV = ρ2 sin φ dρ dθ dφ.
• When doing triple integrals involving the expression
x2 + y 2, I will think “cylindrical coordinates”.
• When doing triple integrals involving the expression
x2 + y 2 + z 2, I will think “spherical coordinates”.
• I will practice interchanging the order of integration.
• I will practice setting up integrals for the test.
• I will practice problems like Problem (3) on this test.