Multivariate Calculus
Review Problems for Final
Final: Tuesday, September 4, 3:30-5:30 PM, Randell Hall 326
• Emphasis will be on topics not tested during the term: Section 14.3: Double Integrals in Polar
Coordinates, Section 14.4: Parametric Surfaces (tangent planes and surface area); Section 14.5:
Triple Integrals; Section 14.6: Triple Integrals in Cylindrical and Spherical Coordinates; Section
13.8: and also Maxima and Minima for Functions in Two Variables.
• For convenience, here is the list of sections from the course syllabus: 11.1 – 11.7; 12.1 – 12.2;
13.1 – 13.8; 14.1 – 14.7
1. Group One
(1) Find the total mass of the solid that lies in the first octant and is bounded by the sphere
x2 + y 2 + z 2 = 9 with mass density function f (x, y, z) = z 2 .
Comments: Use spherical coordinates. The total mass is given by
! Z
Z π/2
Z π/2 Z π/2 Z 3
3
π
2
4
2 2
cos φ sin φ dφ
ρ dρ
[ρ cos φ] ρ sin φ dρ dφ dθ =
2
0
0
0
0
0
=
81
π 1 35
= π
2 3 5
10
(2) Set up only the triple integral for the mass of the solid that lies in the first octant and
is bounded by the planes x = 0, y = 0, z = 0, and (x/2) + (y/3) + (z/4) = 1. The mass
density function f (x, y, z) = x2 + y 2 + z.
Z 2 Z y=3(1−x/2) Z z=4(1−x/2−y/3)
Comments:
(x2 + y 2 + z) dz dy dx
0
y=0
z=0
(3) Set up only a triple integral in cylindrical coordinates for the total
p mass of the solid
2
2
bounded above by the paraboloid z = 4 − x − y and below by z = 3 x2 + y 2 with mass
density g(x, y, z) = x2 + y.
Z 2π Z 1 Z z=4−r2
Comments:
[r2 cos2 θ + r sin θ] r dz dr dθ
0
0
z=3r
(4) Find the area of the portion of the cone given by
r(u, v) = u cos v i + u sin v j + u k
for which 1 ≤ u ≤ 2, 0 ≤ v ≤ π/2.
Comments:
i
j
k
∂r ∂r
sin v 1 = h−u cos v, −u sin v, ui
×
= cos v
∂u ∂v −u sin v u cos v 0 p
√
√
So kh−u cos v, −u sin v, uik = u2 cos2 v + u2 sin2 v + u2 = 2u2 = 2u. Then the surface
area is given by
√
Z 2 Z π/2 √
√ π 1
π 2
2u dvdu = 2
(4 − 1) =
2 3
2
1
0
1
(5) Find the absolute maximum and minimum values of f (x, y) = xy − x − y on the closed
triangular domain T with vertices (0, 0), (4, 0), and (0, 4).
Comments: Since fx = y − 1 = 0 and fy = x − 1 = 0, there is a unique critical point (1, 1)
that lies inside the triangle and f (1, 1) = −1. To check for an absolute max and min, we
still need to check the values of f on the three line segments that make up the boundary of
the triangle T .
(1) From (0, 0) to (4, 0), we have y = 0 and 0 ≤ x ≤ 4. Then f (x, 0) = −x. Clearly, on this
segment, the maximal value is 0 and minimum value is −4.
(2) From (0, 0) to (0, 4), x = 0 and 0 ≤ y ≤ 4. Then f (0, y) = −y. Clearly, on this segment,
the maximum value again is 0 and minimum value is −4.
(3) From (0, 4) to (4, 0), y = −x + 4 with 0 ≤ x ≤ 4. Then f (x, y) = f (x, −x + 4) =
x(−x+4)−x−(−x+4) = −x2 +4x−4. Set w(x) = −x2 +4x−4. Then w0 (x) = −2x+4 = 0
so x = 2 with w(2) = 0.
We conclude that the absolute maximum is 0 and the absolute minimum is −4.
(6) Find the tangent line to the curve of intersection of the two surfaces −x2 + y 2 + z 2 = 6
and z 2 = x2 + 5y 2 at the point (2, 1, 3).
Comments: Let f (x, y, z) = −x2 + y 2 + z 2 and g(x, y, z) = x2 + 5y 2 − z 2 . Then ∇f =
h2x, 2y, 2zi and ∇g = h2x, 10y, −2zi. At the point P (2, 1, 3), these gradients give the
two
vectors
2h2, 1, 3i and 2h2, 5, −3i. The tangent line is parallel to their cross product
i j k 2 1 3 = h−18, 12, 8i. So the tangent line is given as x = 2 − 9t, y = 1 + 6t, z = 3 + 4t.
2 5 −3
(7) Consider the function f (x, y) = 3x2 + 5y 3 at the point P (1, 1).
(a) Find the directional derivative of f in the direction v = i + 2 j at the point P .
(b) At the point P , find the unit vector that makes the directional derivative have its
largest value. At the point P, find the unit vector that makes the directional derivative
have the value zero.
Comments: (a) ∇f = h6x, 15y 2 i. At the point P (1, 1), we obtain the vector ∇f (P ) =
h6, 15i = 3h2, 5i. Let v√ = i + 2 j, then Dv f (P ) = ∇f (P ) · v/kvk. So the directional
derivative has value 36/ 5.
√
(b) The direction√that gives the largest value is h2, 5i/ 29; gives the value zero in the
direction h−5, 2i/ 29.
(8) (a) Consider the points A(1, 0, 1), B(2, 1, 4), and C(3, 1, 2). Find the area of the triangle
with vertices A, B, C.
(b) Consider the fourth point D(3, 4, 3). Find the volume of the parallelopiped determined
by the edges AB, AC, and AD.
Comments:
(a) AB = h1, 1, 3i, AC = h2, 1, 1i. Their cross product is AB × AC =
i j k
√
√
1 1 3 = h−2, 5, −1i. The area of the triangle is 1 4 + 25 + 1 = 1 30.
2
2
2 1 1 1 1 3 (b) AD = h3, 4, 3i. The volume of the parallelepiped is the triple scalar value 2 1 1 =
3 4 3 11.
(9) Find the point on the graph of z = 2 + xy + y 2 whose tangent plane is orthogonal to the
line x = 5 − t, y = 2 + t, z = 1 + 21 t.
2
Comments: The normal vector N to the tangent plane is hy, x + 2y, −1i. The required
point satisfies N = Ch−1, 1, 1/2i. So C = −2. Hence y = 2 and x = −6.
2. Group Two
(1) Find the parametric equations of the line that passes through the point (6, 2, 0) and is
perpendicular to the plane 4y + 3z = 5.
Comments: A normal vector n to the plane is h0, 4, 3i which will be a parallel vector to
the line. Its parametric equations are x = 6 + 0t = 6, y = 2 + 4t, z = 0 + 3t = 3t.
(2) Find a unit vector in the direction in which f (x, y) = sin(3x − y) increases most rapidly at
(π/4, π/2), and find the rate of change of f at P0 (π/4, π/2) in that direction.
Comments: The partial derivatives of f are fx = 3 cos(3x − y)√and fy = − cos(3x − y).
At the point P0√
, their values are fx (π/4, π/2) =
√ 3 cos(π/4)
√ = 3 2/2 and fy (π/4, π/2) =
− cos(π/4) = − 2/2. Hence ∇f (π/4, π/2) = h3 2/2, − 2/2i.
The unit vector a in the direction in which f increases most rapidly at P0 (π/4, π/2) is
√
√
√
√
√
√
h3 2/2, − 2/2i
h3 2/2, − 2/2i
h3 2/2, − 2/2i
3
1
√
√
√
=p
=
= h √ , − √ i.
kh3 2/2, − 2/2ik
5
10
10
(18/4) + (2/4)
√
The maximal rate of change in this direction is k∇f (P0 )k = 5.
(3) Consider the parallelepiped with adjacent edges
u = h1, −1, −1i,
v = h0, 2, 1i,
w = h1, 0, 2i.
(a) Find the area of the face determined by v and w.
(b) Find the volume of the parallelepiped determined by u, v and w.
Comments: (a) The area of this face is kv × wk. But
i j k v × w = 0 2 1 = h4, 1, −2i.
1 0 2 p
√
The area is 42 + 12 + (−2)2 = 21.
(b) The volume is given by the triple scalar product |u · v × w| with value
1 −1 −1 2
1 = h1, −1, −1i · h4, 1, −2i = 5.
u · v × w = 0
1
0
2 (4) Evaluate the integral by reversing the order of integration
Z 4Z 2 p
x3 + 1 dxdy.
√
0
y
Comments: In reversed order, we have the double integral
Z 2 Z x2 p
Z 2 p
3
x + 1 dydx =
x2 x3 + 1 dx.
0
0
0
x3
Make the substitution u =
+ 1 with du = 3x2 dx. Then we find that
Z 2 p
Z
1 9√
1 2 3/2 9 52
2
3
x x + 1 dx =
u du = · u = .
3 1
3 3
9
0
1
3
(5) Find the equation of the tangent plane to the parametric surface x = uv, y = u − v, z = u2
at the point where u = 1 and v = 2.
Comments: We first find the partial derivatives
∂y
∂z
∂x
= v,
= 1,
= 2u.
∂u
∂u
∂u
Write r = r(u, v) = hx(u, v), y(u, v), z(u, v)i. Then
∂r
∂r
= hv, 1, 2ui,
(1, 2) = h2, 1, 2i.
∂u
∂u
Similarly, we have the derivatives relative to v:
∂x
= u,
∂v
∂y
= −1,
∂v
∂z
= 0.
∂v
and
∂r
∂r
= hu, −1, 0i,
(1, 2) = h1, −1, 0i.
∂v
∂v
The normal vector n to the tangent plane is given by
i
j k ∂r
∂r
1 2 = h2, 2, −3i.
×
= h2, 1, 2i × h1, −1, 0i = 2
∂u ∂v
1 −1 0 The point on the plane is given by (u, v) = (1, 2) is
y = u − v = 1 − 2 = −1,
x = uv = (1)(2) = 2,
z = u2 = 1.
Hence the tangent plane is
2(x − 2) + 2(y + 1) − 3(z − 1) = 0, or 2x + 2y − 3z = −1.
(6) Convert the given double integral in rectangular coordinates to an equivalent integral in
polar coordinates. Do not evaluate this integral.
Z 2 Z √4−x2 p
x2 + y 2 dydx.
0
0
Comments: The
p domain of integration D is the quarter circle of radius 2 in the first
quadrant. Also x2 + y 2 = r and dA = rdrdθ so
Z 2 Z √4−x2 p
Z π/2 Z 2
Z π/2 Z 2
2
2
x + y dydx =
r · r drdθ =
r2 drdθ
0
0
0
0
0
2
x +
0
(7) Consider the p
solid G that is bounded above by the sphere
+ z 2 = 16 and below by
the cone z = x2 + y 2 .
(a) Set up only a triple integral in rectangular coordinates for the volume of G.
(b) Set up only a triple integral in cylindrical coordinates that represents the volume of G.
(c) Set up only a triple integral in spherical coordinates that represents the volume of G.
2
Comments: We need to
p find the circle of intersection between the sphere and cone: x +
2
2
2
2
2
2
2
2
y + z = 16 and
√ 2z = 16 or
√ z = x + y . But z = x + y so substituting we find
2
z = 8 so z = 2 2. We conclude that the circle of intersection has radius 2 2.
(a) In rectangular coordinates, the triple integral is
Z 2√2 Z √8−x2 Z √16−x2 −y2
dz dy dx.
√
√
√
−2 2
− 8−x2
x2 +y 2
4
y2
(b) In cylindrical coordinates, the triple integral is
Z 2π Z 2√2 Z √16−r2
r dzdrdθ.
0
0
r
(c) In spherical coordinates, the triple integral is
Z 2π Z π/4 Z 4
ρ2 sin φ dρdφdθ.
0
0
0
(8) Use the transformation u = x + y, v = x − y to find the value of
ZZ
2
2
(x − y)ex −y
D
over the domain D enclosed by the lines x + y = 0, x + y = 1, x − y = 1, and x − y = 4.
Comments: Since u = x + y, v = x − y, we can solve for x, y to get
1
x = (u + v),
2
1
y = (u − v).
2
The Jacobian is given by
∂(x, y) xu xv 1/2 1/2
=
=
∂(u, v) yu yv 1/2 −1/2
= −1
2
In u, v, the domain D in the xy-plane corresponds to 0 ≤ u ≤ 1 and 1 ≤ v ≤ 4. Then the
double integral becomes
ZZ
Z Z
1 4 1 uv
x2 −y 2
(x − y)e
=
ve dudv
2 1 0
D
u=1
Z
Z
1 4 v
1
1 4 uv dv =
e (e − 1) dv = (e4 − e − 3).
=
2
2
2
1
1
u=0
(9) Consider the rectangular box that is open on the top and has a volume of 32 cubic feet.
What should the dimensions of the box be for it to have the minimum surface area?
Comments: Let x be the length, y the width, and z the height of the box. Then its volume
V = xyz. By assumption, V = 32 so z = 32/xy and x > 0, y > 0, z > 0. Its surface area is
S = xy + 2xz + 2yz or
64 64
+ .
S(x, y) = xy +
x
y
To minimize, we find the critical points of S(x, y):
Sx = y −
64
64
= 0 =⇒ y = 2 ,
2
x
x
Sy = x −
64
64
= 0 =⇒ x = 2 .
2
y
y
So x2 y = xy 2 = 64. Canceling, we obtain x = y. Substituting back, we find x = 64/x2 or
x3 = 64. We conclude x = 4 and so y = 4; that is, the critical point is (4, 4).
For the second derivative test, we need
Sxx =
192
,
x3
Syy =
192
,
x3
Sxy = 0.
2 =
At the critical point (4, 4), we find that Sxx = 3 and Syy = 3. Then D = Sxx Syy − Sxy
9 > 0 and Sxx = 3 > 0, so (4, 4) is indeed a minimum.
The dimensions of the box are x = 4, y = 4, and z = 2.
5
3. Group Three
2
2
(1) Consider the
p solid bounded above by the paraboloid z = 10 − x − y and below by the
cone z = 3 x2 + y 2 .
(a) Find the circle of intersection between the paraboloid and cone. (b) Set up the triple
integral in cylindrical coordinates for the volume of the solid.
Comments: In cylindrical coordinates the surfaces have the equations: z = 10 − r2 and
z = 3r. So their intersection is determined by 3r = 10−r2 or r2 +3r−10 = (r+5)(r−2) = 0.
So the circles has the equation r = 2 on the horizontal plane z = 6.
The triple integral is
Z 2π Z 2 Z z=10−r2
r dz dr dθ.
0
0
z=3r
(2) Find the surface area of the portion of the paraboloid z = 4 − x2 − y 2 that lies above the
xy-plane as follows:
(a) Find the differential of surface area dS for the paraboloid. (b) Write down the double
integral for the surface area and evaluate it.
Comments: If z = f (x, y), then the differential dS of surface area is given by
q
dS = 1 + fx2 + fy2 dA(x, y).
p
Since z = f (x, y) = 4 − x2 − y 2 , we find that dS = 1 + 4x2 + 4y 2 dA(x, y).
The area is given by
Z 2π Z 2 p
Z 17
2
1 + 4r r dr dθ = 2π
u1/2 ( 18 du) = π6 (173/2 − 1)
0
0
1
where we used the change of variables u = 1 + r2 with du = 8rdr.
(3) Set-up only the triple integral in spherical coordinates of the mass of the solid that
lies in the first octant bounded between two spheres whose centers are
p at the origin and
whose radii are 3 and 4 and is bounded laterally by the cone z = x2 + y 2 with mass
density function f (x, y, z) = x2 + y 2 . Make sure your answer is written using only spherical
coordinates.
Comments:
Z π/2 Z π/4 Z 4
Z π/2 Z π/4 Z 4
2
2
2
(ρ sin φ) ρ sin φ dρ dφ dθ =
ρ4 sin3 φ dρ dφ dθ
0
0
3
0
0
3
(4) (a) Set-up only the triple integral for the total of the solid that lies in the first octant and
is bounded by the coordinate planes as well as the plane 2x + 3y + z = 6. The mass density
function f (x, y, z) is given by x2 + y 2 + z.
(b) Consider the function f (x, y) = 3x2 + 5y 3 . Find the unit vector a for which the
directional derivative (Df )a (P0 ) at the point P0 = (1, 1) attains its largest value.
Comments: (a)
Z 3 Z y=(6−2x)/3 Z z=6−2x−3y
dz dy dx.
0
(b) ∇f =
h6x, 15y 2 i
y=0
z=0
√
so ∇f (1, 1) = h6, 15i. So the desired vector is a = h2, 5i/ 29.
(5) Consider the parametric surface given by
r(u, v) = uvi + uev j + veu k.
6
(a) Find the partial derivatives of position vector r(u, v) and their values when u = ln 2
and v = 0. (b) Find their cross product. (c) Find the tangent plane to the surface when
u = ln 2 and v = 0.
Comments:
∂r
∂r
= hv, ev , veu i,
= hu, uev , eu i
∂u
∂v
∂r
∂r
At u = ln 2 and v = 0, we get ∂u
= h0, 1, 0i and ∂v
= hln 2, ln 2, 2i with cross product
h2, 0, − ln 2i. Hence the tangent plane has equation 2x − ln 2z = 0 or 2x = ln z.
(6) The company I-M-Pets is designing a pet cage and wants to minimize its cost of construction.
Basically, the cage is a box with no top. The cage will have the volume of 1 cubic foot. The
material of the bottom of the cage has the cost of $4 per square foot, while the material
for the sides of the cage has the cost of $2 per square foot. Find the dimensions of the cage
that will minimize its cost. Assume that x is the length and y is the width of the bottom
while z is the height of the cage. Organize your solution as follows: (a) Express the cost
function C(x, y) of the cage in terms of x and y. (b) Find the critical points of C(x, y).
(c) Apply the Second Derivative Test.
Comments: The volume V = xyz = 1 and the total cost C = 4xy + 4xz + 4yz so we
may take C = xy + xz + yz then multiply this value separately by 4. Then C(x, y) =
xy + 1/x + 1/y.
1
∂C
1
∂C
= y − 2 = 0,
=x− 2 =0
∂x
x
∂y
y
so x2 y = 1 and xy 2 = 1; that is, x2 y = xy 2 which implies x = y since x > 0 and y > 0. But
x2 y = 1 or x3 = 1 so x = 1. It follows y = 1 and z = 1 as well.
Now
∂2C
2
∂2C
2
∂2C
=
,
=
,
= 1.
∂x2
x3
∂y 2
y3
∂y∂x
At x = y = 1, we get fxx = fyy = 2 and fyx = 1. Then ∆ = fxx fyy − fx y 2 = 4 − 1 = 3.
Since fxx > 0, we have a minimum.
(7) (a) Use the chain rule to find the value of ∂f /∂u when u = 1 and v = −1 if
√
f (x, y) = x2 y 3 , x = u, y = uv 2 .
(b) Consider rectangle with length x and width y. Assume that dx/dt = 5, dy/dt = −2. At
the instant when x = 7 and y = 3, at what rate is the diagonal of the rectangle changing?
Comments:
∂f
∂f ∂x ∂f ∂y
=
+
= (2xy 3 ) 12 u−1/2 + (3x2 y 2 ) v 2 .
∂u
∂x ∂u ∂y ∂u
When u = 1 and
p v = −1, we get the value 4.
(b) g(x, y) = x2 + y 2 , then
dg
∂g ∂x ∂g ∂y
x
y
29
=
+
= · 5 + · (−2) = √ .
dt
∂x ∂t
∂y ∂t
g
g
58
(8) (a) Find all points of intersection of the line x = 1 + t, y = 1 − t, z = 1 + t with the cone
z 2 = x2 + y 2 .
(b) Find the plane that contains the line x = −2+3t, y = 4+2t, z = 3−t and is perpendicular
to the plane x − 2y + z = 5.
Comments: (a) Consider (1 + t)2 = (1 + t)2 + (1 − t)2 so we get t = 1 with point (2, 0, 2).
7
(b) Two parallel vectors to the desired plane are v = h3, 2, −1i and w = h1, −2, 1i. Their
cross product h0, −4, 8i is the normal vector to the desired plane which has the equation
(y − 4) + 2(z − 3) = 0,
y + 2z = 10.
4. Group Four
Additional Problems
(1) (a) Find the distance between the two lines L1 : x = 1 + t, y = 2 − t, z = −1 + 2t and
L2 : x = t, y = 1 − t, z = 3t.
(b) Find the distance between the point (1, 0, 1) and the plane 2x − 3y + z = 3.
(2) (a) Find the rate of change of f at the point P in the direction of the vector v where
f (x, y) = xy + 3x2 , P (1, −1), v = 2 i + 3 j.
(3)
(4)
(5)
(6)
(7)
(8)
(b) Find the unit vector in the direction of the maximal rate of change of the function f at
the point P , and find that maximal rate.
Find a point on the surface given by the graph of z = f (x, y) = x2 − 2y 2 where the tangent
plane is parallel to the plane 8x − 4y + z = 3.
√
Consider the helix C given parametrically by x = 2 cos t, y √
= 2 sin t, z = 2t. Find the arc
length of the helix between the points (2, 0, 0) and (0, 2, π/ 2).
(a) Where does the line x = 1 + t, y = 3 − t, z = 2t intersect the cylinder x2 + y 2 = 16?
(b) Find the value of the scalar k so that the vector from the point A(1, −1, 3) to the point
B(3, 0, 5) is orthogonal to the vector from A to the point P (k, k, k).
2
2
2
Find the mass
py + z = 4 and below by the
p of the solid bounded above by the sphere x +
cone z = x2 + y 2 and where the mass density is given by x2 + y 2 + z 2 .
Find the mass of the solid that lies under the paraboloidp
z = x2 + y 2 and above the ring
2
2
9 ≤ x + y ≤ 16 and where the mass density is given by x2 + y 2 .
Convert the iterated integral into polar coordinates and evaluate
Z 2 Z √2x−x2 p
x2 + y 2 dydx.
0
0
(9) Convert the given triple
integral into spherical coordinates:
Z 3 Z √9−x2 Z √9−x2 −y2 p
z x2 + y 2 + z 2 dzdydx,
(a)
√
2
0
−3 − 9−x
Z Z √
Z √
9−y 2
3
(b)
18−x2 −y 2
√
0
0
(x2 + y 2 + z 2 ) dzdxdy.
x2 +y 2
8