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Unformatted text preview: Mathematics 234, Fall 2004 Lecture 3 (Wilson)
Second Midterm Exam November 15, 2004 ANSWERS
Problem 1 (a) Set up but do not evaluate an integral in spherical coordinates to compute the volume of the
part of the sphere of radius 3, centered at the origin, where z is nonnegative. (Le. the region
in the sphere $2 —I y2 —I— Z2 g 9 which is above the xy—plane.) Answer: 42” (Jr/2 /3 P2 Sin(¢) dp dq5 d6 7r 26 4—7‘
(b) Evaluate the triple integral / / / 2T2 dz dr d6
0 0 0 Answer: 20 4 r2 20( 8
[fr/0 /0 2rzdzdrd62/07T/0( 4r—r3 )drd02/0W(862— 464)d6=—7r3——7r5. Problem 2 The volume of a cylindrical can of radius 7' and height h is V(7', h.) = 7TT2 h. A certain can is intended to have a radius of 7' = 2 inches and a height of h = 5 inches. Due to
manufacturing tolerances, the radius changes to 7' = 2.01 inches and the height changes to h = 4.98
inches. Use partial derivatives to approximate how much the volume of the can changes from its designed
value. Answer: VT 2 27rrh, which at (2,5) gives 207T. V1,: 7rr2 ,which at (2,5) gives 47r. The change
in 7" is Ar = 0.01 and the change in h is Ah: —0.02. Thus the change in V is approximately
0.01 X 2071' —I (—0.02) X 47? = 7r(0.2 — 0.08) = 0.1271 (The actual change is 011969871) Problem 3 Find an equation for the tangent plane to the surface 9033—1—3myz—I—2y —z3 —=15 at the point (1, —1,2). Answer: Lett1ngf(ac,y,z—) — 1E 33+3xyz+2y —23 ,we have Vf= (3002 I 33/2)? I (3:172 I 6y2 )jl (330g 322 )k. At (1, —1, 2) that gives —3i’—I— 12f— 151?. Since the vector from (1, —1, 2) to any point in the tangent
plane must be perpendicular to that gradient vector, we can write an equation for the plane through
(1, —1,2) perpendicular to Vf as 3(33 1) I 12(y I 1) 15(z 2) = 0, which can be simpliﬁed to
ac — 43/ + 52 = 15. Problem 4 (a) Compute /// (332 + 22) dV, where R is the region in the ﬁrst octant bounded by
R $2+z2=4andm+y=2 Answer: (There are various orders in which to set this up, here is one example. All should result
in the same numeric answer.) /// W + 23) dV 2 (,2 AH [)me + 22) dz dy dag
R 2 2717 1 2 1
2/ / (——$3—$2+2$+4)dyd$=/ (—x4—4m2+8)d$
0 0 2 0 2 2_128
0 15‘ 0 \/4—m2
(b) Convert the integral / / (:1:2 + 3/2 + 2:13) dy d$ to polar coordinates.
—2 —\/4—m2 You do not have to evaluate this integral. Answer: The region of integration is the half of the circle of radius 2, with center at the origin,
that is on the left of the yaxis. In polar coordinates we get 3—” 2
/2 / r(r2+2rcos(0))drd0.
ﬂ 0 2 (The function f($, y) = $2 + y2 + 2m has some symmetries but not all. Integrating over another
region, rotated about the origin, would not generally give the same answer, and is not an
equivalent answer.) Problem 5 Find and identify all relative maxima, relative minima, and saddle points, for the function f(w,y) =w3 +92 —6w2 +y— 1 Answer: fl. = 3:132 — 12$ and fy 2 231+ 1. Setting those equal to zero we get 3$(:13 — 4) = 0, so :1: = 0 or
90 = 47 and y = —%. There are no singular or boundary points so we examine the points (0, —%) and
(4, —%). Using the second derivative test, fm = 6:10 — 12: At (0, —%) that gives —12 and at (4, —%) it
produces 12. More simply, fyy = 2. And fig 2 0. Thus D = fmfyy — fig is —24 at (0, —%), so that is
a saddle point, and D gives 24 at (4, —%), so that is either a relative maximum or a relative minimum:
Since farm 2 12 > 0, it is a relative minimum. Problem 6 Set up but do not evaluate an integral to compute the surface area for the section of the sphere x2 + y2 + Z2 = 4 above the anyplane and inside the vertical cylinder :32 + (y — 1)2 = 1. Answer: We can take the sphere above the my— plane to be the graph of z 2 v4 — $2 — yg. Then
: — _ #7; fa; \/4——$2——112 and fy— Hence what we need to integrate is 1/4— $2—y2 1E2 y2 4
«/ 2 2 1: : :1: —.
fw+fy+ \/4—x2—y2 4—x2—y2 \/4_$2_y2 The region of integration is the circle in the any—plane with center (0, 1) and radius 1. We can then set
up the integral as 1+\/1— m2 4
/11/ $22dydx.
1—\/1—2:v _ _ 3/2 Problem 7 A thin plate covers the triangular region in the plane whose edges are the ac—axis, the line y = 2m, and
the line :1: = 1.
The density of this plate is given by the function 6(90, 3/) = 6:10 + 63/ + 6. (a) Use an integral to evaluate the mass of this plate. Answer: We integrate 6 (:10, y) through the triangular region, 1 217
mass = / / (6:17 + 63/ + 6) dy dx = 14 mass units, unspeciﬁed.
0 0 (b) Find the moment M5,; of the plate about the $—axis. Answer: 1 2m
Mmz/ / y(6m+6y+6)dydx=11.
0 0 (c) Find the moment My of the plate about the y—axis. Answer:
1 230
My =// a:(6a:+6y+6)dyda: = 10.
0 0 (d) Find the coordinates (mg) of the center of mass of this plate. Answer:
—_ My 10_ 5 Mac _ 11
£3 _ mass 14— —’7 and y— _ mass _ H’ (e) Find the moment of inertia (second moment) of this plate about the :IJaxis7 LU. Answer: 1 2m
1302/ / y2(6$+6y+6)dyd$=12.
0 0 (f) Find the radius of gyration of this plate about the $—axis. Answer: _ Isa _ /E_\/§
— mass_ 14— 7' Problem 8 Find the largest and smallest values of f($, y) = $2 + y subject to $2 + 3/2 = 4. Answer: We use Lagrange Multipliers, seeking points (:10, y) where Vf(ac, y) is parallel to Vg($, y), for
g(:v,y) = x2 + 3/2 — 4. Vf = 2xf+ j’and V9 2 211077+ 2yji Where those are parallel, Vf = AVg for
some constant A. That makes 2$T+ f: 2$AT+ 2ij’ so 2$ = 2$A and 1 2 23M. From the ﬁrst of
those equations we have 2:1:(1 — A) = 0 so either x— — 0 or A— — 1. If A— — 1 the second equation gives 2y=10ry=%. Soeither:1:=00ry=1 Usingtheequationg($, y)=0, i.e. $2 +3;2 =4, if$=0 then y: ::2, while if y = l we have 510—— :‘/—. (ﬂ 1) and< m 1) 2 72 _27§' We can evaluate f at each of these four points: f(0,2) = 2, f(0, 2) = 2, f(21,2)= 41, and
f(—£, l)2 4%. Comparing the results, we see the largest value for f is 41, taken at both (4L, ;)
and (—9, %). while the smallest value is —2 which is taken only at (0, —2). 5 Thus we have four points to consider, (0,2), (0, —2), ...
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 Multivariable Calculus

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