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18.02 Final Review Problems
Last Updated: Dec. 18, 2017
These are selected solutions to the final review problems. Please e-mail me (Jane – [email protected])if you find mistakes or if there is a solution that you want to see that I haven’t written yet.
1 Vectors, Matrices, Parametric equations
1. Consider the two planes x+ y = 1 and y + 2z = 3.
(a) * Find the angle of intersection between these two planes.
(b) ** Find a parametrized equation for the line of intersection of the two planes.
2. ** Does there exist a plane that contains both the lines (x, y, z) = (t − 1, 2t, 2t − 5) and(x, y, z) = (2t, t+ 8, 3t− 1). If so, find an equation for this plane.
3. ** Find the distance between the point (1, 0, 0) and the plane x+ 2y − z = 0.
4. *** Find the distance between the lines given by (x, y, z) = (t, 2 − t, 1 + 2t) and (x, y, z) =(1− t, t, 1).
It’s tempting to try to use the distance formula and minimize d2 = (t− (1− t))2 + ((2− t)−t)2 + ((1 + 2t)− 1)2. But this doesn’t work, since really we should be parametrizing the twolines with different variables (x, y, z) = (t, 2− t, 1 + 2t) and (x, y, z) = (1− s, s, 1).
To find the distance between the two lines, we need to either do a minimization problem overtwo variables, or we need to find a vector v that is orthogonal to both lines, and a vectorw that goes between two points on the two lines, and find the length of the projection of wonto v. Let’s use the second method.
A vector orthogonal to both lines is v = (1,−1, 2)× (−1, 1, 0) = (−2,−2, 0). Substituting int = 0 in both lines gives points (0, 2, 1) and (1, 0, 1) on the two lines. A vector between themw = (0, 2, 1)− (1, 0, 1) = (−1, 2, 0). Then,
projvw =v ·w‖v‖2
v =−2
8(−2,−2, 0) = (1/2, 1/2, 0).
This vector has length 1/√
2, which is then the minimum distance. .
5. Consider the parametrized curve x(t) = (cos(3t), sin(3t), 2t3/2).
(a) ** Sketch this curve for 0 ≤ t ≤ 2π. Calculate the velocity and acceleration as functionsof t.
(b) * Find the length of this curve from t = 0 to t = 2π.
(c) * Calculate the unit tangent vector and the unit normal vector as a function of time.
(d) ** Find the curvature and radius of curvature as a function of time.
1
2 Partial derivatives, Tangent planes, Min-Max problems
1. ** Consider the sphere of radius 3 centered at the origin. The plane tangent to the sphere at(1, 2, 2) intersects the x axis at the point P . Find the coordinates of P .
The tangent plane at the point (1, 2, 2) has normal vector (1, 2, 2) and therefore has equation(x− 1) + 2(y − 2) + 2(z − 2) = 0. When y = z = 0, we have that x = 0. Therefore, the pointis (9, 0, 0).
2. ** Suppose that you have a function f(x, y) where the directional derivative at the point (2, 2)in the direction (1/
√2, 1/√
2) is 5 and the directional derivative in the direction (−1/√
2, 1/√
2)is −1. Find the gradient of f at (2, 2). Find the directional derivative in the the direction(1, 3).
3. * Find the gradient of f(x, y) = sin(x)ey at (π/2, 0). Use this to approximate the value of(sin(π/2− 0.1)e−0.1).
4. ** For which point(s) on the curve x2 + 2xy+ z = 3 is the tangent plane parallel to the planex+ 2y + 3z = 0?
The normal vector to the tangent plane of a point on the curve is (2x+ 2y, 2x, 1). We wantthis normal vector to be parallel to (1, 2, 3), and therefore
c(2x+ 2y, 2x, 1) = (1, 2, 3)
for some constant c. Equating the z coordinates gives us that c = 3. So we have that6x+6y = 1, 6x = 2, giving us that x = 1/3, y = −1/6. Our desired point is then (1/3,−1/6, 3).
5. The temperature as a function of location is T (x, y) = 2x2 + 3y2.
(a) * Find the direction of fastest increase for temperature at the point (1, 2).
The gradient of T is ∇T = (4x, 6y). Therefore, at (1, 2), the direction of fastest increaseis in the direction of ∇T (1, 2) = (4, 12), which is (1/
√10, 3/
√10).
(b) * Find an approximation for the temperature at the point (1.01, 1.98).
A formula for the approximation is
z − z0 = 4(x− x0) + 12(y − y0).
For x0 = 1, y0 = 2, z0 = 14, this gives z = 14 + 4(−0.01),+12(0.02) = 14.2.
(c) ** Suppose that you want to walk always in the direction of fastest increase of temper-ature. Find a curve describing the path that you want to walk, going through the point(1, 2).
We want a curve C(t) = (x(t), y(t)) that goes through (1, 2) and has that C ′(t) =(4x(t), 6y(t)), the gradient at (x(t), y(t)). This gives us the system of differential equa-tions {
dxdt = 4xdydt = 6y.
Solving these gives x(t) = Ae4t, y(t) = Be6t. Letting A = 1, B = 2 so that the curvepasses through (1, 2) at time t = 0 gives us our final answer of C(t) = (e4t, 2e6t).
2
(d) *** Suppose that you will walk along the circle x2 + y2 = 1 at a constant speed. Atwhat points along your walk will the temperature be increasing the fastest?
This problem is asking for the time at which your directional derivative in the direction oftravel is the largest. If we parametrize our curve as C(t) = (cos(t), sin(t)), our directionof travel at time t is (− sin(t), cos(t)). Dotting this with the gradient at time t gives usour directional derivative
(4 cos(t), 6 sin(t)) · (− sin(t), cos(t)) = 2 sin(t) cos(t) = sin(2t).
This is maximized at t = π/4 and t = 5π/4. So the points of our walk at which thetemperature is increasing the fastest are (
√2/2,√
2/2) and (−√
2/2,−√
2/2).
6. ** Describe the set of points in space for which the tangent planes of the graph of f(x, y) =x2 − y2 + 1 and the surface g(x, y, z) = xy − z2 are perpendicular.
7. *** Find the min and max value of the function f(x, y) = xy on the set x2 + y2 + xy ≤ 1.
On the interior, we find where the gradient is zero and use the second derivative test.∇f(x, y) = (y, x) is zero only at the point (0, 0). The second derivative test then tells usthat this is a saddle point since D(x, y) = −1 < 0. This point cannot be a minimum ormaximum.
Now we use Lagrange multipliers to find minima/maxima on the boundary. We want to findpoints where the gradient of
L(x, y, λ) = xy − λ(x2 + y2 + xy − 1)
is zero. This gives us the system of linear equationsy − 2λx− λy = 0
x− 2λy − λx = 0
x2 + y2 + xy − 1 = 0.
The first two equations give us that
y
y + 2x=
x
x+ 2y,
so xy + 2y2 = xy + 2x2, giving us that x = ±y. If x = y, then the third equation is 3x2 = 1,so the possible points are (±1/
√3,±1/
√3). If x = −y, then the third equation is x2 = 1, so
the possible points are (±1,∓1).
We have that
f((±1/√
3,±1/√
3)) = 1/3
f((±1,∓1)) = −1.
So the minimum of −1 is attained at the points (±1,∓1) and the maximum of 1/3 is attainedat the points (±1/
√3,±1/
√3).
8. *** Find the absolute min and max values on the closed triangular region with vertices(0, 0), (2, 0), and (0, 2) of the function f(x, y) = 2x2 − 2xy + y2 − y + 3.
3
We need to find the critical points of the function f on the interior of the triangle as well ason the boundary.
On the interior, we have that ∇f(x, y) = (4x − 2y,−2x + 2y − 1). This has its only zero at(−1/2,−1), which is not in the triangle.
The boundary is broken up into three pieces, the three sides of the triangle. They can beparametrized by
C1(t) = (t, 0)
C2(t) = (0, t)
C3(t) = (t, 2− t).
Each of these curves go from t = 0 to t = 2. We can substitute each of these parametrizationsinto f(x, y) and find critical points of f on these curves:
f(C1(t)) = 2t2 + 3
f(C2(t)) = t2 − t+ 3
f(C3(t)) = 5t2 − 7t+ 5.
We can then check by finding where the derivatives of these three functions are zero thatf(C1(t)) has a critical point at t = 0, which is at (0, 0). f(C2(t)) has a critical point at t = 1/2,which is at (0, 1/2). f(C3(t)) has a critical point at t = 7/10, which is at (7/10, 13/10). Sothe points that could be the minimal and maxima are at these critical points and also theendpoints of the line segments (0, 0), (2, 0), (0, 2). We can compute the value of f on thesepoint:
point value
(0, 0) 3(2, 0) 11(0, 2) 5
(0, 1/2) 11/4(7/10, 13/10) 51/20
Therefore, the absolute min value is 51/20 at (7/10, 13/10) and the absolute max value is 11at (2, 0).
9. ** Find the critical points of the function f(x, y) = x4 − 2xy2 + y8 on R2. Find the minimalvalue of this function and where it is attained.
The critical points occur where
(4x3 − 2y2,−4xy + 8y7) = (0, 0).
One solutions is (0, 0), which has a function value of 0. Away from (0, 0), the second equationgives us that x = 2y6, and substituting in to the first equation gives us that 16y16 = 1.Therefore, y = ±2−1/4 and x = 2−1/2. Therefore, we have two more critical points at(2−1/2,±2−1/4). The value of f at both of these points is −1/2, so −1/2 is the minimum andis attained at these points. [We notice that f(x, y) = x4 − 2xy2 + y8 ≥ x4 − x2 + y8 − y4,which is large and positive if (x, y) is far from the origin, so this is a true minimum.]
4
10. *** Find and classify all critical points of the function f(x, y) = 3x2y + 6xy2 + y3. (Theremight be a point where the second derivative test is inconclusive. What can you do here?)
The gradient is∇f(x, y) = (6xy + 6y2, 3x2 + 12xy + 3y2).
The first component of this is 6xy+ 6y2 = 6y(x+ y), which is zero when y = 0 or x = −y. Ineither of these cases, checking when the second component is also zero gives that the gradientis zero at (0, 0).
The matrix of second partials is [6y 6x+ 12y
6x+ 12y 12x+ 6y
].
At (0, 0), the determinant of this matrix is 0, and therefore the second derivative test isinconclusive. We notice though that f(0, 0) = 0, and fixing x = 0 and letting y vary near 0gives that f(0, y) > 0 for small positive y and f(0, y) < 0 for small negative y. Therefore, fmust have neither a local min or max at the origin, and (0, 0) is a saddle point.
11. ** A cylindrical metal can is to be manufactured from a fixed amount of sheet metal. Usethe method of Lagrange multipliers to determine the ratio between the dimensions of the canwith the largest capacity.
12. *** Integrate the function x2y2 over the region x2 + y2 ≤ 1.
Using polar coordinates, we have that∫ ∫x2+y2≤1
x2y2 dx dy =
∫ 2π
0
∫ 1
0r5 cos2 θ sin2 θ dr dθ
=1
6
∫ 2π
0cos2 θ sin2 θ dθ
=1
6
∫ 2π
0
1− cos2 2θ
4dθ
=1
6
∫ 2π
0
1− cos 4θ
8dθ
=π
24.
13. ** Find the total mass of an object with density function ρ(x, y) =√x2 + y2 over the circle
of radius 1 centered at (1, 0).
14. ** The radius of a cone is increasing at a rate of 2 while the height is decreasing at a rate of1.5. At what rate is the volume of the cone changing when the radius and height are 10 and12 respectively? Is the volume increasing or decreasing?
5
We have that V = 13πr
2h. Therefore, using the chain rule, we have that
dV
dt=dV
dh
dh
dt+dV
dr
dr
dt
=
(1
3πr2)dh
dt+
(2
3πrh
)dr
dt
=
(1
3π · 102
)(−1.5) +
(2
3π(10)(12)
)(2)
= 110π.
The volume of the cone is increasing at a rate of 110π.
15. * The function f is in terms of the variables u and v. We also know that u and v are functionsof x and y, where u = x+ y, v = x− 2y. Find df
dx in terms of f , u, and v.
3 2D: Line integrals, Area integrals, Green’s theorem
1. Consider the function f(x, y) = x− y2.
(a) ** Find a vector field F such that the curl of F is f .
Answer: For example, let F = (y3/3, x2/2).
(b) *** Use Green’s theorem to rewrite∫∫R
(x− y2) dA
where R is the elliptical region 4x2 + y2 ≤ 4 as some line integral over the boundary ofR. Then compute this line integral.
Let R denote the elliptical region, and C be its boundary, oriented counterclockwise. ByGreen’s theorem, we have that∫∫
R(x− y2) dA =
∮C
y3
3dx+
x2
2dy.
Parametrizing C by C(t) = (cos(t), 2 sin(t)), we have that our integral is∮C
y3
3dx+
x2
2dy =
∫ 2π
0
8 sin3(t)
3(− sin(t))dt+
cos2(t)
2(2 cos(t)) dt
= −8
3
∫ 2π
0sin4(t) dt+
∫ 2π
0cos3(t) dt.
By symmetry, we have that the latter integral is 0. Using the power reduction formula,we have that
−8
3
∫ 2π
0sin4(t) dt = −8
3
∫ 2π
0
(1− cos(2t))2
4dt
= −2
3
∫ 2π
0(1− 2 cos(2t) + cos2(2t)) dt
= −4π
3+ 0− 2
3· 1
2
∫ 2π
01 dt.
= −2π
6
2. Let F be the vector field given by F(x, y) = 2xi + 3y2j.
(a) * Show that F is a conservative field and find a potential function.
We can check that F(x, y) is conservative since My = Nx = 0 and that a potential isf(x, y) = x2 + y3.
(b) *** What is the largest amount of work that can be done by F over a curve C thatremains inside a square with vertices at (1, 1), (−1, 1), (−1,−1), and (1,−1)?
The work done by F over a curve from the point (a, b) to the point (c, d) is
f(c, d)− f(a, b) = c2 + d3 − a2 − b3.
With the constraint that |a|, |b| ≤ 1, this is maximized when c2 = d3 = 1, b3 = −1, anda2 = 0. Therefore, the largest amount of work that can be done is 3.
(c) ** Compute the work done by F around the curve S, the perimeter of the square withvertices in part (b), oriented counterclockwise.
Let R denote the region inside the square. Since the curl Nx−My = 0, Green’s theoremtells us that ∮
S2x dx+ 3y2 dy =
∫∫R
0 dA = 0.
3. Consider the integral∫ 10
∫ 1√x sin(y3) dy dx.
(a) * Sketch the region of integration.
The region of integration is the triangular region in the below picture:
(b) ** Evaluate the integral.
The inner integral is difficult to evaluate as is. This is a clue that we might want tochange the order of integration. Doing this, we get that∫ 1
0
∫ 1
√x
sin(y3) dy dx =
∫ 1
0
∫ y2
0sin(y3) dx dy
=
∫ 1
0y2 sin(y3) dy
=
[−cos(y3)
3
]10
=1
3− cos(1)
3.
7
4. ** Evaluate the integral∫ 10
∫ 1y x
2 sin(xy) dx dy.
This integral is difficult to evaluate as is. Let’s try to change the order of integration. Todo this, it helps to first draw the region of integration (the triangular region in the belowpicture):
Then, changing the order of integration gives us∫ 1
0
∫ 1
yx2 sin(xy) dx dy =
∫ 1
0
∫ x
0x2 sin(xy) dy dx
=
∫ 1
0[−x cos(xy)]x0 dx
=
∫ 1
0(−x cos(x2) + x) dx
=
[− sin(x2)
2+x2
2
]10
=− sin(1)
2+
1
2.
5. ** Find the work done by F = (x2y,−x − y) around the curve C that is a trapezoid withvertices (0, 0), (3, 0), (3, 1), and (1, 1).
6. ** (Problem Edited, Dec. 14) Evaluate the integral∫ 3
2
∫ 4−x
−2+x
x− y(x+ y)3
dy dx.
The integrand looks unpleasnt in this problem, which is a suggestion to do a change ofvariables. Let’s try u = x−y and v = x+y. If we draw out the original region of integration,it is a triangle with vertices at (2, 0), (2, 2), and (3, 1), bounded by the lines x = 2, y = 4− x,and y = −2 + x.
By using the substitution x = u+v2 and y = v−u
2 , we see that the new region of integrationin u and v coordinates is a triangle with vertices at (2, 2), (0, 4), and (2, 4) bounded by the
8
curves u+ v = 4, v = 4, and u = 2. We can calculate the Jacobian to be J = 1/2. Therefore,we have that ∫ 3
2
∫ 4−x
−2+x
x− y(x+ y)3
dy dx =
∫ 4
2
∫ 2
4−v
u
2v3du dv
=
∫ 4
2
[u2
4v3
]24−v
dv
=
∫ 4
2
(−3v−3 + 2v−2 − v−1/4
)dv
=
[3v−2
2− 2v−1 − ln(v)
4
]42
=7− 8 ln(2)
32.
7. ** Let F = (2xy,−y2 − 2). Let C be a curve in the first quadrant that starts at (a, 0), endsat (0, b), and does not intersect itself. Find the flux of F through C.
8. *** Let F =(−y
x2+y2, xx2+y2
). Let C be a curve in the first quadrant that starts at (a, 0), ends
at (0, b), and does not intersect itself. Find the work done by F over C.
Since we are not given the curve C and the vector field has a singularity, we should try to useGreen’s theorem. To do so, we need to close the curve so that it bounds some region. Thework done by F is easy to calculate around circular arcs centered at the orign and straightline segments extending radially from the orign. We can assume without loss of generalitythat a < b, and close C with a circular arc A from (0, b) to (b, 0) and a straight line B from(b, 0) to (a, 0):
If we let R be the region bounded by these three curves, we have that∫CF · dr +
∫AF · dr +
∫BF · dr =
∫∫R
div(F) dx dy.
We can calculate the divergence of F to be zero, so the right hand side is equal to zero. Thevector field F is tangent to A everywhere, and orthogonal to B everywhere. This gives us
that∫B F · dr = 0. On A, the unit tangent at a point (x, y) is
(y
x2+y2, −xx2+y2
). Therefore,
9
F · T = 1/(x2 + y2) = −1/b2 on A. The length of the segment A is π · b2/2, so∫AF · dr =
∫AF · T ds =
∫A
−1
b2ds =
−length(A)
b2=−πb2/4b2
=−π4.
Putting this all together, we get that∫CF · dr =
∫∫R
div(F) dx dy −∫AF · dr−
∫BF · dr = 0−
(−π4
)− 0 =
π
4.
9. * (Problem Edited, Dec. 14) Find the mass of a rope that follows the curve y = x2 + 1from the point (0, 1) to (2, 5), with density ρ(x, y) = x.
The integral that we need to evaluate is ∫Cx ds,
where C is the curve. Parametrizing the curve C(t) = (t, t2 + 1) from t = 0 to 2, we havethat ds =
√(x′(t))2 + (y′(t))2 dt =
√1 + 4t2 dt. Then, our integral becomes∫
Cx ds =
∫ 2
0t√
1 + 4t2 dt
=
[(1 + 4t2)3/2
12
]20
=17√
17− 1
12.
4 3D: Line integrals, Volume integrals, Divergence theorem, Stokes’theorem
1. ** Find the volume of the solid that is the intersection of two right cylinders x2 + y2 ≤ a andy2 + z2 ≤ a.
The integral to evaluate is∫ a
−a
∫ √1−y2
−√a2−y2
2√a2 − y2 dx dy =
∫ a
−a4(a2 − y2) dy = 16a3/3.
2. Suppose that a solid sphere of radius 2 centered at the origin intersects a solid vertical cylinderof radius 1 centered at (1, 0, 0) and height 4 in the z direction (−2 ≤ z ≤ 2).
(a) ** Find the volume of the intersection.
The most straightforward way to do this problem is probably via cylindrical coordinates.Our r and θ bounds then sweep out a circle of radius 1 centered at (1, 0) in the xy-plane,
10
and the z bound goes from −√
4− x2 − y2 = −√
4− r2 to√
4− r2. Computing thisintegral, we get that
Volume =
∫ π
0
∫ 2 cos θ
0
∫ √4−r2−√4−r2
r dr dθ
=
∫ π
0
∫ 2 cos θ
02√
4− r2r dr dθ
=
∫ π
0
[−2
3(4− r2)3/2
]2 cos θ0
dθ
=
∫ π
0
−2
3
((4− 4 cos2 θ)3/2 − 8
)dθ
=16
3
∫ π
0(1− sin3 θ) dθ
=16
3
∫ π
0(1− sin θ + cos2 θ sin θ) dθ
=16
3
[θ + cos θ − cos3 θ
3
]π0
=16
3π − 64
9.
(b) *** Find the surface area of the top part of the intersection.
We’re going to do this problem by projecting the top part of the surface down to the xy-plane. Let’s call the shadow region R. Here, the surface is the graph z =
√4− x2 − y2 =
f(x, y), and the region that it projects to is the circle of radius 1 centered at (1, 0). Wealso have that
fx =−x√
4− x2 − y2, fy =
−y√4− x2 − y2
.
Then, our integral can be computed as:
Surface Area =
∫∫R
√1 + f2x + f2y dy dx
=
∫∫R
√1 +
x2
4− x2 − y2+
y2
4− x2 − y2dy dx
=
∫ π
0
∫ 2 cos θ
0
2√4− r2
r dr dθ
=
∫ π
0
[−2(4− r2)1/2
]2 cos θ0
dθ
=
∫ π
0(−2(4− 4 cos2 θ)1/2) + 4) dθ
=
∫ π
0(−2(4 sin2 θ)1/2 + 4) dθ
=
∫ π
0(−4 sin θ + 4) dθ
= [4 cos θ + 4θ]π0= 4π − 8.
11
Notice that in the fourth to last step, we have a (4 sin2 θ)1/2 term. This turns into|2 sin θ|, which is equal to 2 sin θ for θ between 0 and π. If we had instead used thebounds −π/2 to π/2 for θ, we would have to split the integral into two parts. For θbetween −π/2 to 0, we replace |2 sin θ| with −2 sin θ. For θ between 0 and π/2, wereplace |2 sin θ| with 2 sin θ.
3. ** Compute the mass of the first octant of a unit sphere with density ρ(x, y, z) = x+ y + z.
We set up and compute this integral in spherical coordinates, making the appropriate substi-tutions for x, y, and z:
Volume =
∫ π/2
0
∫ π/2
0
∫ 1
0(ρ sinϕ cos θ + ρ sinϕ sin θ + ρ cosϕ)ρ2 sinϕdρ dϕdθ
=
∫ π/2
0
∫ π/2
0
∫ 1
0ρ3(sin2 ϕ(cos θ + sin θ) + cosϕ sinϕ)dρ dϕdθ
=1
4
∫ π/2
0
∫ π/2
0(sin2 ϕ(cos θ + sin θ) + cosϕ sinϕ)dϕ dθ
=1
4
∫ π/2
0
(π
4(cos θ + sin θ) +
[sin2 ϕ
2
]π/20
)dθ
=1
4
∫ π/2
0
(π
4(cos θ + sin θ) +
1
2
)dθ
=1
4
[π
4(sin θ − cos θ) +
1
2θ
]π/20
=3π
16.
4. ** Let S be defined by z = e1−x2−y2 for z ≥ 1, oriented by the upward pointing normal. Let
F = (x, y, 2− 2z). Calculate∫∫S F · dS.
This integral looks difficult to compute directly, so let’s try to close up the surface andapply the divergence theorem. Let S be the original surface and T be the unit disk atx2 + y2 ≤ 1, z = 1, with unit normal oriented downward. Then, since div(F) = 0, we havethat ∫∫
SF · dS +
∫∫TF · dS = 0.
On T , the unit normal is (0, 0,−1), so F · n = (x, y, 2− 2z) · (0, 0,−1) = 2z− 2, which is zeroon the disk since z = 1 here. Therefore,
∫∫T F · dS = 0, so
∫∫S F · dS = 0.
5. Let V be an inverted solid cone with a vertex at the origin, a flat circular base at z = 2,constant density 1, and a cone angle of π/3.
(a) ** Find the moment of inertia of the cone around the z axis.
Let’s do this integral in cylindrical coordinates. First, we need the equation of the cone.We know that it must be of the form z =
√c(x2 + y2). In spherical coordinates
z = ρ cosϕ = ρ cos(π/6) = ρ√
3/2
and √c(x2 + y2) =
√cρ2 sin2 ϕ =
√cρ sin(π/6) =
√cρ/2.
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So this means that c = 3, and our equation is z =√c(x2 + y2) =
√3r. As z goes from
0 to 2, r goes from 0 to 2/√
3. The distance to the z-axis of a point (r, θ, z) is r.
Now, our moment of inertia integral is∫∫∫V
(dist)2 · 1 dV =
∫ 2π
0
∫ 2/√3
0
∫ 2
√3rr2 · r dz dr dθ
=
∫ 2π
0
∫ 2/√3
0(2r3 −
√3r4) dr dθ
=
∫ 2π
0
(8
9− 32
45
)dθ
=16π
45.
(b) *** Find the moment of inertia of the cone around the axis vertical axis given by x = 1,and y = 0.
The setup is similar to the previous problem, except that now, the distance squared tothe axis of a point (x, y, z) is (x− 1)2 + y2 = x2− 2x+ 1 + y2 = r2− 2r cos θ+ 1. So ourintegral is∫∫∫
V(dist)2 · 1 dV =
∫ 2π
0
∫ 2/√3
0
∫ 2
√3r
(r2 − 2r cos θ + 1)r dz dr dθ
=
∫ 2π
0
∫ 2/√3
0(−√
3r4 + 2r3 + 2√
3r3 cos θ − 4r2 cos θ −√
3r2 + 2r) dr dθ
=
∫ 2π
0
(28
45− 8 cos θ
9√
3
)dθ
=56π
45
6. *** Let F = (x2y,−xy2− y, z) and let S be a box with unit side lengths, sitting with its basesomewhere on the xy-plane, and missing its top face (i.e. the surface is five sides of a unitcube). Find the flux of F through S, if the normal is pointed outward. Your answer shouldnot depend on where the box sits on the xy-plane. (Hint: try using the divergence threoremby “completing” the box).
Let S be the box missing the top face, and T be the top face. We’ll use the divergencetheorem to solve this problem. We see that div(F) = 2xy − 2xy − 1 + 1 = 0. Therefore,∫∫
SF · dS +
∫∫TF · dS =
∫∫∫V
div(F) dV = 0.
Also, on T , we have that the unit normal is (0, 0, 1), so F · n = z = 1 on this face. Therefore,since the area of the face is 1, we have that
∫∫T F · dS = 1. Putting this all together, we see
that∫∫S F · dS = −1.
7. *** Find the graviational attraction that a point mass M at the point (0, 1, 0) exerts on theupper hemisphere of a solid sphere with center at the origin and radius 1 (x2 + y2 + z2 ≤ 1and z ≥ 0) and constant density ρ.
8. *** Consider a solid sphere of radius 1. Let P be a bound on the surface of the sphere. Whatis the average distance of points inside the solid sphere to the point P?
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9. *** Find the volume of the solid that is the intersection of three right cylinders x2 + y2 ≤ a,y2 + z2 ≤ a, and x2 + z2 ≤ a.
10. *** Let S be a sphere centered at the origin with radius a. If c < d are two numbers between−a and a, find the surface area of a sphere between the heights z = c and z = d. (Youranswer should depend on c, d, and the radius a).
11. *** Let S be the surface z = 2 − x2 − y2, lying above the unit circle x2 + y2 ≤ 1 in thexy-plane. Calculate the flux of
F =
(−4x+
x2 + y2 − 1
1 + 3y2, 3y, 7− z − 2xz
1 + 3y2
)through S. (Hint: you may want to use the divergence theorem.)
This flux looks difficult to calculate directly. We would like to close up the surface and applythe divergence theorem. A possible choice is to close up the surface with a disk x2 + y2 ≤1, z = 1. The normal on this disk is (0, 0,−1), which would make
F · n = −7 + z +2xz
1 + 3y2= −7 + 1 +
2z
1 + 3y2.
This is still difficult to integrate over the disk. So let’s try another choice. Instead of havingthis disk at height z = 1, let’s place it at z = 0 to make F · n = 7, which is much nicer. Then,we also need a cylindrical piece x2 + y2 = 1, 0 ≤ z ≤ 1 to close off our region.
Let V be the 3D region. Then, it is bounded by S (z = 2− x2 − y2, x2 + y2 ≤ 1) our originalsurface at the top, a cylindrical side T (x2 + y2 = 1, 0 ≤ z ≤ 1), and a circular base B(x2 + y2 ≤ 1, z = 0).
We have then that∫∫SF · n dS +
∫∫TF · n dS +
∫∫BF · n dS =
∫∫∫V
div(F) dV.
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For the right hand side, the divergence can be calculated to be div(F) = 2. Then, usingcylindrical coordinates, we have that∫∫∫
Vdiv(F) dV =
∫ 2π
0
∫ 1
0
∫ 2−r2
0(−2)r dz dr dθ
=
∫ 2π
0
∫ 1
0(−4r + 2r3) dr dθ
=
∫ 2π
0(−3/2) dθ
= −3π.
We then need to calculate the flux through the pieces T and B. First for B, we have thatF · n = −7, since z = 0 on B and n = (0, 0,−1). Then,∫∫
BF · n dS =
∫∫B
7 dS = −7Area(B) = −7π.
On the cylindrical piece T , the unit normal at a point (x, y, z) = 1√x2+y2
(x, y, 0) = (x, y, 0)
since x2 + y2 = 1 on this cylinder. Then, we have that for (x, y, z) on the cylinder,
F·n =
(−4x+
x2 + y2 − 1
1 + 3y2, 3y, 7− z − 2xz
1 + 3y2
)·(x, y, 0) =
(−4x, 3y, 7− z − 2xz
1 + 3y2
)·(x, y, 0)
since x2 + y2 = 1 on T . Then, Fn = −4x2 + 3y2.
Then, in cylindrical coordinates, we have that
∫∫TF · n dS =
∫ 2π
0
∫ 1
0(−4 cos2 θ + 3 sin2 theta) · 1 dr dθ
=
∫ 2π
0(−4 cos2 θ + 3 sin2 θ) dθ
= −π,
since∫ 2π0 cos2 θ dθ =
∫ 2π0 sin2 θ dθ = 1
2
∫ 2π0 dθ = π.
Putting this all together, we have that
∫∫SF · n dS =
∫∫∫V
div(F) dV −∫∫
TF · n dS −
∫∫BF · n dS = −3π − (−π)− (−7π) = 5π.
(Note: this problem comes from http://www.math.harvard.edu/archive/21a spring 03/exams.html)
12. ** (Edited, Dec. 18) Find the volume of the region between a paraboloid with equationz = (x− 1)2 + 4y2, and the plane 2x+ z = 5.
Let’s first find the curve of intersection of these two surfaces. We have that they intersect at
(x− 1)2 + 4y2 = 5− 2x,
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which simplifies tox2 + 4y2 = 4.
Letting x = 2r cos θ, y = r sin θ, and z = z, we have that the volume that we want is∫ 2π
0
∫ 1
0
∫ 5−2x
(x−1)2+4y22r dr dθ.
(The factor of 2 comes from the Jacobian in the change of coordinates.) We can then computethis integral as∫ 2π
0
∫ 1
0
∫ 5−2x
(x−1)2+4y22r dr dθ =
∫ 2π
0
∫ 1
02r(5− 2x− x2 + 2x− 1− 4y2) dr dθ
=
∫ 2π
0
∫ 1
02r(−4r2 + 4) dr dθ
=
∫ 2π
0
[−2r4 + 4r2
]10
= 4π.
13. ** Let S be the surface defined by y = 10−x2−z2 with y ≥ 1 and rightward pointing normal.Let
F = (2xyz + 5z, ex cos(yz), x2y).
Determine∫∫S ∇× F · dS.
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