Multivariate Calculus: Review Problems for Examination Two
Note: Exam Two is on Tuesday, August 16. The coverage is multivariate differential calculus and double
integration. You should review the double integration problems handed out last week.
1. Group One
(1) Suppose that the portion of a tree that is usable for lumber is a right circular cylinder. If the usable
height of a tree increases 3 feet per year and the usable radius increases 1 foot per year, how fast is
the volume of usable lumber increasing when the usable height of the tree is 25 feet and the usable
radius is 2 feet?
∂V dr
∂V dh
dV
∂V
∂V
dV
=
+
, or
=
+3
which equals
Comments: Let V = πr2 h. Then
dt
∂r dt
∂h dt
dt
∂r
∂h
2
2πrh + 3 · πr = 100π + 12π = 112π.
(2) (a) Given that the directional derivative of f (x, y, z) at the point (1, 2, 1) in the direction of a =
2i − j − 2k is −5 and that k∇f (1, 2, 1)k = 5, find ∇f (1, 2, 1). (b) Let f (x, y, z) = xy + 3yz. Find
the largest and smallest values of the directional derivative of f at the point (2, 1, 1).
Comments: (a) ∇f (1, 2, 1) = −5a/kak = −(5/3) (2i − j − 2k).
√
√
(b) ∇f = hy, x + 3z, 3yi so ∇f (1, 2, 1) = h2, 4, 6i which has magnitude 22 + 42 + 62 = 2 14.
Recall that the largest value of the directional derivative is the norm of the gradient and occurs in
its direction while the smallest value is the negative of the norm of the gradient which occurs in the
opposite direction.
(3) (a) Find a normal vector to the surface given by the ellipsoid 3x2 + 2y 2 + z 2 = 6 at the point
(1, 1, −1). (b) Find parametric equations for the tangent line to the curve of intersection of the two
surfaces given by 3x2 + 2y 2 + z 2 = 6 and 3xy + 2yz = 1 at the point (1, 1, −1).
Comments: (a) Let f (x, y, z) = 3x2 + 2y 2 + z 2 . The desired normal vector is n = ∇f (1, 1, −1) but
∇f = h6x, 4y, 2zi so n = h3, 2, −1i.
(b) A parallel vector t to the tangent line is given by the cross product of the gradients of the
functions f from part (a) and g(x, y, z) = 3xy + 2yz. Let n2 = ∇g(1, 1, −1) so ∇g = h3y, 3x + 2z, 2yi
so n2 = h3, 1, 2i. Then t = n1 × n2 = h5, −9, −3i. The parametric equations for the tangent line are
x = 1 + 5t, y = 1 − 5t, z = −1 − 3t.
(4) Find a point on the surface z = 16 − 4x2 − y 2 at which the tangent plane is perpendicular to the
line x = 3 + 4t, y = 2t, and z = 2 − t.
Comments: The desired point occurs where the gradient of g(x, y, z) = 16 − 4x2 − y 2 − z is a
multiple of the parallel vector v of the given line so v = h4, 2, −1i. But ∇g = h−8, −2y, −1i. We
need h−8x, −2y, −1i = C h4, 2, −1i. This occurs with C = 1 and so −8x = 4 or x = −1/2, −2y = 2
or y = −1.
(5) Find all saddle points of the z = x2 − 2y 2 − 6x + 8y + 3. Justify your answer.
Comments: The critical point occurs when ∂z/∂x = ∂z/∂y = 0. Now ∂z/∂x = 2x − 6 = 0 so x = 3
and ∂z/∂y = −4y + 8 = 0 so y = 2. The second order partials are ∂ 2 z/∂x2 = 2, ∂ 2 z/∂y 2 = −4,
∂ 2 z/∂x2 ∂ 2 z/∂x∂y
while ∂ 2 z/∂x∂y = 0. Consequently det 2
= −8 so the critical point is a
∂ z/∂x∂y ∂ 2 z/∂y 2
saddle point.
(6) Find the volume of the solid that lies in the first octant and is bounded by the cylinder x2 + z 2 = 4
and the planes z = 0, y = 0, and y = x.
ZZ p
Comments: The volume is given by a double integral
4 − x2 dA where D is the domain in
D
the xy-plane bounded by the lines x = 0, y = 0, y = x, and x = 2. So
ZZ p
Z 2Z
4 − x2 dA =
D
0
=
1
2
Z
x
p
4 − x2 dydx =
0
4
Z
x
0
u1/2 du =
0
1
8
3
2
p
4 − x2 dx
(7) Evaluate the following double integral by reversing the order of integration:
Z 2Z 1
2
e−x dx dy.
y/2
0
Z 1
Z 1 Z y=2x
Z 2 Z x=1
2
2 1
−x2
−x2
2xe−x dx = − e−x = 1 − 1/e
e
dy dx =
e
dx dy =
Comments:
0
x=y/2
0
y=0
0
0
2
(8) Find the volumep
of the solid that is bounded above by the paraboloid z = 6 − (x + y 2 ) and below
by the cone z = x2 + y 2 .
Comments: Use polar coordinates. The surfaces become z = 6 − r2 and z = r. Their circle of
intersection is found as follows: 6 − r2 = r or r2 + r − 6 = (r + 3)(r − 2). So this circle has radius 2.
Then the double integral for the volume is
Z 2π Z 2
[(6 − r2 ) − r] r dr dθ.
0
0
2. Group Two
(1) The volume of a right circular cone of radius r and height h is V = (1/3)πr2 h. Show that if the
∂V
2V
height remains constant while the radius changes, then the volume satisfies
=
.
∂r
r
Comments: We first compute
∂V
∂ 1 2
2
=
πr h = πrh.
∂r
∂r 3
3
On the other hand,
2V
2
=
r
r
1 2
2
πr h = πrh,
3
3
2V
so the equation ∂V
∂r = r holds.
(2) Let T = x2 y − xy 3 + 2; x = r cos θ, y = r sin θ. Find the partials ∂T /∂r and ∂T /∂θ.
Comments: By the chain rule, we have that
∂T
∂r
=
∂T ∂x ∂T ∂y
+
∂x ∂r
∂y ∂r
2x − y 3 cos θ + x2 − 3xy 2 sin θ
2r cos θ − r3 sin3 θ cos θ + r2 cos2 θ − 3r3 cos θ sin2 θ sin θ
=
2r cos2 θ − r3 cos θ sin3 θ + r2 cos2 θ sin θ − 3r3 cos θ sin3 θ.
=
=
Similarly, by the chain rule,
∂T
∂θ
∂T ∂x ∂T ∂y
+
∂x ∂θ
∂y ∂θ
= − 2x − y 3 r sin θ + x2 − 3xy 2 r cos θ
= − 2r cos θ − r3 sin3 θ r sin θ + r2 cos2 θ − 3r3 cos θ sin2 θ r cos θ
= − 2r2 cos θ sin θ − r4 sin4 θ + r3 cos3 θ − 3r4 cos2 sin2 θ
=
(3) Let z = ln(x2 + 1), where x = r cos θ. Find ∂z/∂r and ∂z/∂θ.
Comments: Use the chain rule:
2r cos θ
−2r2 cos θ sin θ
∂z
dz ∂x
2x
=
= 2
(−r sin θ) = 2
(−r sin θ) =
.
2
∂θ
dx ∂θ
x +1
r cos +1
r2 cos2 +1
∂z/∂r is found similarly.
(4) The length, width, and height of a rectangular box are increasing at the rates of 1 in/sec, 2 in/sec,
and 3 in/sec, respectively.
(a) At what rate is the volume increasing when the length x is 2 in, the width y if 3 in, and the
height z is 6 in?
(b) At what rate is the length of the diagonal increasing at that instant?
2
(c) At what rate is the surface area of the box increasing at that instant?
Comments: (a) The volume V = xyz with V (2, 3, 6) = 36. So by the chain rule
dV
dt
∂V dx ∂V dy ∂V dz
+
+
∂x dt
∂y dt
∂z dt
= yz(1) + xz(2) + yz(3)
=
=
18 + 24 + 54 = 96.
p
√
(b)
The
diagonal
of
the
box
is
given
by
f
(x,
y,
z)
=
x2 + y 2 + z 2 so f (2, 3, 6) = 4 + 9 + 36 =
√
49 = 7. Note that
∂f
x
x
=p
=
2
2
2
∂x
f
(x,
y, z)
x +y +z
and, similarly, ∂f /∂y = y/f (x, y, z) and ∂f /∂z = z/f (x, y, z) So by the chain rule:
df
dt
∂f dx ∂f dy ∂f dz
+
+
∂x dt
∂y dt
∂z dt
∂f
∂f
∂f
(1) +
(2) +
(3)
∂x
∂y
∂z
x
y
z
(1) +
(2) +
(3)
f (x, y, z)
f (x, y, z)
f (x, y, z)
2
3
6
2 + 6 + 18
26
(1) + (2) + (3) =
=
.
7
7
7
7
7
=
=
=
=
(c) The surface area S is given by S(x, y, z) = 2(xy + xz + yz) so S(2, 3, 6) = 2(6 + 12 + 18) = 64.
By the chain rule,
dS
dt
=
∂S dx ∂S dy ∂S dz
+
+
∂x dt
∂y dt
∂z dt
2(y + z)(1) + 2(x + z)(2) + 2(x + y)(3)
=
2(9) + 4(8) + 6(5) = 80.
=
(5) Let z = f (u) and u = g(x, y). Express the partial derivative ∂z/∂x in terms of dz/du, ∂u/∂x, and
∂u/∂y.
Comments: By the chain rule, we have
∂z
dz ∂u
=
,
∂x
du ∂x
∂z
dz ∂u
=
∂y
du ∂y
x
(a) Find the directional derivative of the function f at P0 (1, 0) in the direction of
(6) Let f (x, y) = x+y
the point Q(−1, −1).
(b) Find a unit vector u for which Du f (2, 3) = 0.
Comments: (a) The directional derivative of f at P0 in the direction of the vector a is given by
Da f (P0 ) = ∇f (P0 ) ·
a
kak
Now we find that
∂f ∂f
1
x
x
,
i=h
−
,−
i.
∂x ∂y
x + y (x + y)2
(x + y)2
√
√
So ∇f (1, 0) = h0, −1i and a = P0 Q = h−2, −1i with kak = 5. So Da f (P0 ) = 1/ 5.
(b) For Dv f (P0 ) to be zero, we need to find a vector v which is perpendicular to ∇f (1, 0) = h0, −1i.
We can simply take v = i.
(7) Let z = 3x2 − y 2 . Find all points at which p
k∇zk = 6.
Comments: ∇z = h6x, −2yi so k∇zk = 36x2 + 4y 2 = 6 will give an ellipse with the equation
x2 + (y/9)2 = 1.
∇f = h
3
p
(8) Show that the surfaces z = f (x, y) = x2 + y 2 and z = g(x, y) = (1/10)(x2 + y 2 ) + 5/2 intersect at
the point (3, 4, 5) and have a common
√ tangent plane at that point.
Comments: Note that f (3, 4) = 32 + 42 = 5 and g(3, 4) = 25/10 + 5/2 = 5 as well so the two
surfaces do indeed intersect at the point (3, 4, 5). Let n1 be the normal vector to the first surface and
n2 the second. Then the surfaces will share the same tangent plane provided their normal vectors
are multiplies of each
p /∂x, −∂f /∂y, 1i and n2 = h−∂g/∂x, −∂g/∂y, 1i
p other. Recall that n1 = h−∂f
Now ∂f /∂x = x/ x2 + y 2 and ∂f /∂y = y/ x2 + y 2 so at the point (3, 4) ∂f /∂x = 3/5 and
∂f /∂y = 4/5 and n1 = h−3/5, −4/5, 1i. Next, ∂g/∂x = x/5 and ∂g/∂y = y/5 so at the point (3, 4)
∂g/∂x = 3/5 and ∂g/∂y = 4/5 and n2 = h−3/5, −4/5, 1i. Since the normal vectors are identical,
the tangent planes agree.
(9) Find all points on the surface f (x, y, z) = x2 + y 2 − z 2 = 1 at which the normal line is parallel to
the line through the points P (1, −2, 1) and Q(4, 0, −1).
Comments: A normal vector to the level surface f (x, y, z) = x2 + y 2 − z 2 = 1 is given by its
gradient ∇f = h2x, 2y, −2zi. We need to find all points where ∇f is proportional to the vector
v = P Q = h3, 2, −2i; that is, h2x, 2y, −2zi = ch3, 2, −2i where c is a scalar. Hence, we have the
three equations:
2x = 3c, 2y = 2c, −2z = −2c.
It follows immediately that y = c and z = c so y = z and x = 3c/2. Substituting back into the level
surface equation, we find that (3c/2)2 + c2 − c2 = 1 or 3c/2 = ±1 or c = ±2/3. Hence there are two
points are the level surface (1, 2/3, 2/3) and (−1, −2/3, −2/3).
p
(10) Find parametric equations for the tangent line to the curve of intersection of the cone z = x2 + y 2
and the plane x + 2y + 2z = 20 at the point P0 (1, −1, 2). [Hint: the desired tangent line is the line
of intersection of two tangent planes.]
Comments: A tangent vector t to the curve of intersection is given by the cross product of the n1
that is normal to the cone with n2 that is normal to the plane. Now
p
p
n1 = h−∂z/∂x, −∂z/∂y, 1i = h−x/ x2 + y 2 , −y/ x2 + y 2 , 1i
√
√
√
and n2 √
= h1, 2, 2i. At the point P0 (1, −1, 2), n1 = h−1/ 2, 1/ 2, 1i. Now n1 × n2 = h 2 − 2, 1 +
√
2, −3 2/2i. Hence the tangent plane is given by
√
√
√
2 − 2(x − 1) + (1 + 2)(y + 1) − 3 2(z − 2)/2 = 0.
(11) Find all saddle points of the z = f (x, y) = x2 + xy + y 2 − 3x. If there are none, verify this as well.
Comments: First, find the critical points of f (x, y); that is, ∂z/∂x = zx = 2x + y − 3 = 0 and
∂z/∂y = zy = x + 2y = 0. That is, 2x + y = 3 and x + 2y = 0 which has the unique solution
2
is negative at the critical point.
(x, y) = (2, −1). Second, we need to determine if D = zxx zyy − zxy
Now zxx = 2 and zyy = 2 while zxy = 1. Since D = 3 > 0, there are no saddle points. In fact,
(2, −1) is a relative minimum since zxx > 0.
3. Group Three
(1) (a) Set up only the double integral that gives the volume of the solid that lies in the first octant
and is bounded above by the cylinder x2 + z 2 = 16, below by the plane z = 0, and finally laterally
by the two planes x = 4 and y = 2x. Do not evaluate the integral!
Comments: (a) The domain D of integration in the xy-plane is a triangle bounded by the lines
x = 0, x = 4, and y = 2x so its vertices are (0, 0), (4, 0), and (4, 8). So the double integral is
Z 4 Z y=2x p
16 − x2 dy dx.
0
y=0
Z 8 Z y=4
2
(2) (a) Given the double integral
ey dy dx, rewrite it by reversing the order of integration.
0
y=x/2
(b) Evaluate this integral. Show your work for credit.
Comments: (a) The domain D of integration is a triangle in the xy-plane bounded by x = 0,
y = 4, and y = x/2 so its vertices are (0, 0), (0, 4), and (8, 4). Then the integral in reversed order is
4
Z
4
0
x=2y
Z
2
ey dx dy.
Z x=0
4 Z
x=2y
0
Z
get
0
16
4
2
2y ey dy. Make the change of variables u = y 2 with du = 2y dy to
dx dy =
0
x=0
u=16
u
u
= e16 − 1.
e du = e e
(b)
Z
y2
u=0
(3) Find parametric equations for the tangent line to the curve of intersection of the elliptic paraboloid
z = x2 + 3y 2 and the plane x + y − z = −2 at the point P0 (1, 1, 4).
∂z
∂z
Comments: The normal vector N1 to the paraboloid is given by h− ∂x
, − ∂y
, 1i = h−2x, −6y, 1i.
At x = 1, y = 1, N1 = h−2, −6, 1i. The normal vector N2 to the plane is h1, 1, −1i. Their cross
product N1 × N2 = h5, −1, 4i. Hence the tangent line is x = 1 + 5t, y = 1 − t, z = 4 + 4t.
(4) (a) Suppose that a function z = f (x, y) is differentiable at the point P0 (1, 3). Suppose that f (P0 ) = 1,
∂f
∂f
(P0 ) = 2, and
(P0 ) = −1. Use a linear approximation with differentials to estimate the value
∂x
∂y
f (1.02, 2.97).
(b) Given the function z = f (x, y) = 3x + y 2 and that x = u2 − v 2 , y = 2uv, use the chain rule to
∂z
.
find the partial derivative
∂u
Comments: (a) The linear approximation is given by
∂f
∂f
(P0 ) ∆x +
(P0 ) ∆y
∂x
∂y
f (P ) ' 4 + 3 · (0.03) + (−2) · (−0.05) = 4 + 0.12 + 0.10 = 4.19.
f (P ) ' f (P0 ) +
(b)
∂z
∂u
∂z ∂x ∂z ∂y
+
∂x ∂u ∂y ∂u
= y (2u) + (x − 2y) (2v) = 2uv · 2u + (u2 − v 2 − 4uv) 2v
=
=
6u2 v − 2v 3 − 8uv 2
(5) (a) Find all the critical points of the function
1
1
f (x, y) = xy − x2 − y 3 .
2
3
(b) Classify each critical point as a relative maximum, relative minimum, or saddle point.
Comments: (a) fx = y − x = 0 and fy = x − y 2 so x = y and x = y 2 so y = y 2 . Hence y = 0 or
y = 1 so there are two critical points (0, 0) and (1, 1).
2
(b) We need fxx = −1, fyy = −2y, and fxy = 1. So the Hessian ∆ = fxx fyy − fxy
= 2y − 1. At
(0, 0), ∆ = −1 so it is a saddle point. At (1, 1), ∆ = 1 and fxx = −1, so it is a relative maximum.
(6) Let g(x, y, z) = y 2 cos(xz), a = −2i − j + 2k, and P0 (π/2, −1, 1).
(a) Find the value of the directional derivative of g at the point P0 in the direction of the vector a.
(b) Find the maximal value of the directional derivative of g at the point P0 .
Comments: (a) ∇g = h−y 2 z sin(xz), 2y cos(xz), −xy 2 sin(xz)i so ∇g(π/2, −1, 1) = h−1, 0, −π/2i.
a
So Da f (P0 ) = ∇g(P0 ) · kak
= h−1, 0, −π/2i · h−2, −1, 2i/3 = (2 − π)/3.
p
(b) k∇g(P0 )k = | h−1, 0, −π/2ik = 1 + π 2 /4.
4. Group Four
(1) (a) Find the volume of the solid that lies in the first octant and is bounded above by the cylinder
x2 + z 2 = 1 and the planes z = 0, y = 0, and y = x.
(b) Evaluate the following double integral by reversing the order of integration:
Z π/2 Z π/2
sin(x2 ) dx dy.
0
y
5
Comments: (a) The domain of integration is the triangle with vertices (0, 0), (1, 0), and (1, 1). So
we have
ZZ p
Z 1 Z xp
Z 1 p
Z 1
1 − x2 dA =
1 − x2 dydx =
x 1 − x2 dx = 12
u1/2 du = 1/3.
0
0
0
D
(b)
Z π/2 Z
π/2
sin(x2 ) dx dy =
y
Z
0
π/2
y=x
Z
sin(x2 ) dydx =
0
0
0
Z
π/2
x sin(x2 ) dx =
0
1
2
Z
π 2 /4
sin(u) du.
0
This integrates to (1 − cos(π 2 /4))/2.
(2) Identify all the critical points of the function
f (x, y) = xy − x3 − y 2 .
Classify each critical point as a relative maximum, relative minimum, or saddle point.
Comments: Step 1: Find the critical points: fx = y − 3x2 and fy = x − 2y. So we get the system
of equations y = 3x2 and x = 2y. This yields x = 2 · 3x2 or x = 0 and x = 1/6. There are two
critical points: (0, 0) and (1/6, 1/12).
2
Step 2: classifcation: We need fxx = −6x, fyy = −2, and fxy = 1. So D = fxx fyy − fxy
= 12x − 1.
At x = 0, D = −1, so we get a saddle point. At x = 1/6, D = 2 − 1 = 1 > 0. Since fxx < 0, we get
a relative maximum.
(3) (a) Suppose that a function z = f (x, y) is differentiable at the point P0 (2, 1). Suppose that f (P0 ) = 3,
fx (P0 ) = 2, and fy (P0 ) = −1. Use a linear approximation with differentials to estimate the value
f (1.98, 1.01).
(b) Find the point on the surface given by the graph of z = x2 + y 2 at which the tangent plane is
perpendicular to the line x = 1 + 3t, y = 2 + 4t, z = −1 + 2t.
Comments: (a)
f (P ) ' f (P0 ) + fx (P0 )∆x + fy (P0 )∆y = 3 + 2(−0.02 + (−1)(0.01) = 2.95.
(b) The surface normal n = h2x, 2y, −1i while the parallel vector v to the line is v = h3, 4, 2i. We
need that n = cv where c is a scalar. We obtain
h2x, 2y, −1i = c h3, 4, 2i,
so c = −1/2. Consequently, x = −3/4 and y = −1 to yield the point (−3/4, −1, 25/16).
(4) Find parametric equations for the tangent line to the curve of intersection of the paraboloid z =
x2 + y 2 and the plane 2x + y − z = −1 at the point P0 (1, 2, 5).
Comments: The normal to the surface at P0 is n1 = h2x, 2y, −1i = h2, 4, −1i while n2 = h2, 1, −1i
is normal to the plane. Their cross product provides the tangent vector
i j k 2 4 −1 = h−3, 0, −6i.
2 1 −1 We take t = h1, 0, 2i as the tangent vector to give the line
x = 1 + t,
2
y = 2,
z = 5 + 2t.
2
(5) (a) Let z = f (x, y) = x − 3y , and x = r cos θ, y = r sin θ. Use the chain rule to find the partial
derivative ∂z/∂θ.
(b) Consider a circular cylinder with radius r and height h. If the radius is increasing at the rate of
2 meters per second and the height is increasing by the rate of 3 meters per second, find the rate is
the volume changing when the radius is 5 meters and the height is 1 meter using the chain rule.
Comments: (a) The chain rule yields
∂z
∂θ
=
=
∂z ∂x ∂z ∂y
+
∂x ∂θ
∂y ∂θ
2x(−r sin θ) − 6y(r cos θ)
= −2r2 cos θ sin θ − 6r2 cos θ sin θ = −8r2 cos θ sin θ.
6
(b) Since V = πr2 h and dr/dt = 2 and dh/dt = 3, we find that
∂V dr ∂V dh
dr
dh
dV
=
+
= 2πrh + πr2
= 20π + 75π = 95π.
dt
∂r dt
∂h dt
dt
dt
5. Double Integral Review Questions
(1) Sketch the domain of integration of the following double integrals:
Z 1 Z y=1
Z 1 Z y=x
(x + y 2 ) dydx,
(x + y 2 ) dydx, (b)
(a)
0
1
Z
y=0
x=y
Z
(x + y 2 ) dxdy,
(c)
0
0
1
Z
y=x
x=1
Z
(x + y 2 ) dxdy.
(d)
x=0
x=y
0
RR
√
(2) Evaluate the double integral D xy dA where D is the domain bounded by the curves y = x,
y = 6 − x, and y = 0. Comments: If we write dA = dydx, we √
need two double integrals to perform
x at the point (4, 2). To see this,
the evaluation.
Note
that
the
line
y
=
6
−
x
intersects
y
=
√
solve x = 6 − x or x = (6 − x)2 which reduces to x2 − 13x + 36 = 0. This quadratic factorizes as
(x − 4)(x − 9). Since 9 lies outside the domain, the point must be (4, 2). Hence we find that
ZZ
Z 4 Z y=√x
Z 6 Z y=6−x
xy dA =
xy dydx +
xy dydx.
D
0
y=0
4
y=0
If we write dA = dxdy, then there is only one double integral
ZZ
Z 2 Z x=6−y
xy dA =
xy dxdy.
D
x=y 2
0
RR
(3) FindRRthe values ofRRthe following where D is the domainRR
enclosed by the unit circle: RR(a) D 5dA,
(b) D x dA, (c) DRR(3x − 4y). RR
Comments: (a) Since D dA = Area(D), we have D 5dA = 5π.
(b,c) By
RR symmetry, D x dA = D y dA = 0. So the answer for both parts is 0.
(4) Find D ydA where D is the domain in the xy-plane bounded between the line x + y = 5 and the
circle x2 + y 2 = 25. Comments: The line and circle have two intersection points at (5, 0) and
(0, 5). So we need to compute
ZZ
Z 5 Z y=√25−x2
y dy dx
y dA =
D
y=5−x
0
y=√25−x2
1 2 dx
y
2 y=5−x
5
Z
=
0
=
1
2
Z
5
(25 − x2 ) − (5 − x)2 dx
0
which is an integral of a polynomial in x.
(5) Find the volume of the solid in the first octant bounded above by the paraboloid z = x2 + 3y 2 and
below by the plane z = 0, and laterally by y = x2 and y = x. Comments:The base D of the solid
in the xy-plane is bounded between the curves y = x2 and y = x. We find that
ZZ
Z 1 Z y=x
(x2 + 3y 2 ) dA =
(x2 + 3y 2 ) dydx
D
y=x2
0
Z
1
=
0
Z
=
0
which is easy to evaluate.
7
y=x
(x2 y + y 3 ) y=x2 dx
1
2x3 − (x4 + x6 ) dx
(6) Find the volume of the solid that is common to the cylinders x2 + y 2 = 25 and x2 + z 2 = 25 that
lies above the xy-plane Comments: The base D of the solid is just the domain bounded by the
circle x2 + y 2 = 25 so the volume is given by
Z 5 Z y=√25−x2 p
ZZ p
2
25 − x dA =
25 − x2 dy dx
√
y=− 25−x2
−5
5
D
y=√25−x2
p
25 − x2 y dx
√
2
Z
=
y=− 25−x
−5
5
Z
=
(25 − x2 ) dx
2
−5
which is easy to evaluate.
(7) Set up only the double integral for the volume of the solid bounded above by the paraboloid z =
x2 + y 2 , below
by the xy-plane, and laterally by the cylinder x2 + (y − 1)2 = 1. Comments:
Z 1 Z y=1+√1−x2
(x2 + y 2 ) dy dx.
√
−1
y=1− 1−x2
(8) Find the volume of the solid in the first octant bounded above by the graph of z = 9 − x2 , below
by z = 0, and laterally by y 2 = 3x. Comments: The domain in the xy-plane is bounded between
the parabola y 2 = 3x and the line obtained from the intersection of z = 9 − x2 with z = 0 so x = 3.
Then the volume is given by
ZZ
Z 3 Z x=3
(9 − x2 ) dA =
(9 − x2 ) dxdy
D
x=y 2 /3
0
Z
3
=
0
Z
=
x=3
(9x − x3 /3)x=y2 /3 dy
3
[(27 − 9) − (3y 2 − y 6 /34 )] dy
0
which is an easy integral to evaluate in y. In reverse order, we get the equivalent double integral
ZZ
Z 3 Z y=√3x
(9 − x2 ) dA =
(9 − x2 ) dy dx
D
y=0
0
3
Z
=
0
3
Z
=
y=√3x
(9 − x2 )y y=0 dx
√
(9 − x2 )(− 3x) dx
0
=
√ Z 3 √
− 3
(9 x − x5/2 ) dx
0
which again is an straightforward integral to evaluate.
Z
(9) Reverse the order of integration in the following double integrals: (a)
Z 4 Z x=8
Z 1 Z x=ey
(b)
f (x, y) dxdy, (c)
f (x, y) dxdy. Comments:
0
x=2y
Z
2Z
0
x=4
(a)
0
8
Z
y=x/2
(b)
x=y 2
Z
f (x, y) dydx,
0
Z πZ
Z
√
y= x
f (x, y) dydx,
0
y=0
x=1
Z
f (x, y) dxdy,
4
y=0
Z
y=1
(c)
f (x, y) dydx.
1
π
e
y=ln x
sin x
dx dy. (a) Describe the domain in the xy-plane given by
x
0
y
the limits of integration. This domain is given in terms of “horizontal cross-sections.” (b) Write
down the equivalent double integral by reversing the order of integration by giving the domain by
(10) Consider the double integral
8
means of “vertical cross-sections.” (c) EvaluateZ theZ integral in part (b).
Z π
y=x
π
sin x
sin x dx = 2.
Comments: The integral in reversed order is
dy dx =
x
y=0
0
0
9