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Math 241, Spring 09
Practice Problems for Exam 4
1. Let R denote the region of points (x, y) ∈ R^2 satisfying x^2 + y^2 ≤ 2 and x ≥ 1.
(a) Rewrite the following integral in polar coordinates:
I =
(x,y)∈R
F (x, y) dA =
θ=...
r=...
F (... ,.. .)...
(b) Evaluate the integral from (a) for F (x, y) = x. Hint:
d dt (tan^ t) = 1/^ cos
(^2) t.
(c) Find the center of mass (¯x, y¯) of the region R, assuming density 1.
2. Let D denote the tetrahedron with the vertices (0, 0 , 0), (1, 0 , 0), (0, 2 , 0), (0, 0 , 3).
(a) Find
(x,y,z)∈D
x dV
(b) Find the center of mass (¯x, y,¯ ¯z) of the region D, assuming density 1.
3. Let D denote the region of points (x, y, z) ∈ R^3 satisfying x^2 + y^2 + z^2 ≤ 2 and z ≥
x^2 + y^2.
(a) Find the volume of D using cylindrical coordinates.
(b) Find the volume of D using spherical coordinates.
4. Let R denote the region of points (x, y) ∈ R^2 satisfying 2 x−^1 ≤ y ≤ 3 x−^1 and (1 − x)−^1 ≤ y ≤ 2(1 − x)−^1. Find
the integral
(x,y)∈R
y
2 dA. Hint: Use the variables u = xy and v = (1 − x)y.
5. Consider the cylindrical surface S consisting of points (x, y, z) ∈ R^3 satisfying x^2 + z^2 = 4.
(a) Let S 1 be the part of S with − 4 ≤ y ≤ 4. Find a parametrization of S 1 and use it to compute the surface area.
(b) Let S 2 be the part of S with 0 ≤ z ≤ y and y ≤ 4. Find a parametrization of S 2 and use it to compute the surface area.
Solutions see next page
Solutions
See also Matlab solutions which have plots for each problem
(a) bounds for θ: x = r cos θ ≥ 1 , r ≤
2 imply cos θ ≥ 1 /
2 ⇐⇒ θ ∈ [−π/ 4 , π/4]
x = r cos θ ≥ 1 ⇐⇒ r ≥ 1 / cos θ, bounds for r: 1 / cos θ ≤ r ≤
I =
∫ (^) π/ 4
θ=−π/ 4
r=1/ cos θ
F (r cos θ, r sin θ)r dr dθ
(b) M 1 =
∫ (^) π/ 4
θ=−π/ 4
r=1/ cos θ
r cos θr dr dθ,
r=1/ cos θ r
(^2) cos θ dr = cos θ
[
1 3 r
3
]√ 2
1 / cos θ =^
2 3
2 cos θ − 13 (cos θ)−^2
M 1 =
[ 2
3
2 sin θ − 13 tan θ
]π/ 4 θ=−π/ 4
(c) mass M =
∫ (^) π/ 4
θ=−π/ 4
r=1/ cos θ
r dr dθ =
π
2
− 1 , M 2 =
∫ (^) π/ 4
θ=−π/ 4
r=1/ cos θ
r sin θr dr dθ = 0 since integrand is
antisymmetric with respect to θ. Hence center of mass is x¯ = M 1 /M = 23 /
π 2 −^1
, y¯ = M 2 /M = 0.
(a) M 1 =
x=
∫ (^2) − 2 x
y=
∫ (^3) − 3 x−(3/2)y
z=
x dz dy dx =
(b) mass M =
x=
∫ (^2) − 2 x y=
∫ (^3) − 3 x−(3/2)y z=0 1 dz dy dx=^1 ,^ M^2 =^
x=
∫ (^2) − 2 x y=
∫ (^3) − 3 x−(3/2)y z=0 y dz dy dx^ =^
1 2 ,^ M^3 = ∫ (^1) x=
∫ (^2) − 2 x y=
∫ (^3) − 3 x−(3/2)y z=0 z dz dy dx^ =^
3 4 , center of mass:^ x¯^ =^ M^1 /M^ =^
1 4 ,^ y¯^ =^ M^2 /M^ =^
1 2 ,^ z¯^ =^ M^3 /M^ =^
3 4
(a) V =
∫ (^2) π
θ=
r=
∫ √ 2 −r 2
z=r
r dz dr dθ =
π
(b) V =
∫ (^2) π
θ=
∫ (^) π/ 4
φ=
ρ=
ρ^2 sin φ dρ dφ dθ =
π
4. Solve xy = u and (1 − x)y = v for x, y: y = u + v, x = u/(u + v)
Find Jacobian determinant:
det
[ (^) ∂x ∂u ,^
∂y ∂u ∂x ∂v ,^
∂y ∂v
]
= det
[
v (u+v)^2
−u (u+v)^2 ,^1
]
u + v
∫ ∫
(x,y)∈R
y
2 dA =
u=
v=
(u + v)
u + v
dv du = 4
(a) parametrization: x = 2 cos u, y = v, z = 2 sin u, 0 ≤ u ≤ 2 π, − 4 ≤ v ≤ 4 ru × rv = (−2 sin u, 0 , 2 cos u) × (0, 1 , 0) = (−2 cos u, 0 , −2 sin u), ‖ru × rv‖ = 2
∫ (^2) π
u=
v=− 4
2 dv du = 32π
(b) parametrization: x = 2 cos u, y = v, z = 2 sin u, 0 ≤ u ≤ π, 2 sin u ≤ v ≤ 4
∫ (^) π
u=
v=2 sin u
2 dv du = 8(π − 1)
