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Unformatted text preview: Mathematics 2347 Fall 2005 Lecture 2 (Wilson)
Second Midterm Exam November 17, 2005 ANSWERS
Problem 1 (a) Let f (x, y) = x2 + 2y2. Set up7 but do not evaluate, an integral to compute the surface
area on the graph of f over the region R which is the triangle with vertices (07 0), (0, 2)7
and (1, 0). ANSWER: The triangle has for its boundaries the lines 36 = 0, y = 0, and y = 2 — 236. One way to
set up the integrals would be 1 27290
/ / ...dy dm.
0 o For surface area we want the integral of ,/ fl? + f; + 1. In this case f3; = 235 and fy 2 43/.
Hence we can use to calculate the surface area 1 27290
/ / V4362 + 16y2 dy dx.
0 0 (b) Set up the following integral with the order of integration reversed, i.e. as an integral
with dy dm replacing dm dy. Do not evaluate the integral! 7 /12 /jﬂ(e$ cos(y)) dm dy 18 ANSWER: At the right is a sketch of the region of integration. As a dy dx
integral, we see the overall range of x values is from 0 to 1. For any x
in that range7 y goes from the lower line y = 1 up to the curved upper
boundary. The curve is x = y — 1 or y = $2 + 1. Hence the integral in
this form is 18 E 8
GB 1 $2+1
/ / (ex cos(y)) dy dx. 0 1 " 0.2 '3" 0.2 034 as BB 1 Problem 2 Set up an iterated integral to compute /// (x + 2y) dV,
R where R is the region in space bounded by the cylinder 362 + 22 = 4, the plane 3/ = 0, and the
plane 3/ + 2 = 2. You did not have to evaluate this integral, but you were offered 5 extra—credit points for correctly
evaluating it. ANSWER: The cylinder has radius 2 and the y—aXis as its center line. The plane y+ z = 2 cuts
across it7 slanting down (2 = 2 — y) as you go out in the y direction. The widest part of the (ellipse) where the plane and cylinder meet, in the x direction, is in the x—y plane where z = 0:
Picture the circle x2 + 22 = 4 in the plane y = O, and x extends from —2 to 2. If we put dx on
the outside, this will give the range for the corresponding integral. If we put dz next, moving
inward, the same circle shows that 2 ranges from —\/4 — $2 to x/ 4 — .762. Now for any .76 and 2
corresponding to a point in that circle, the range of y values is from y = 0 at the x—z plane out
to the angled plane y = 2 — 2. Thus we can set up the integral as 2 x/4ix2 272 d d d
2 .
[Mam/0 (96+ 9) y 2 96 Evaluating this integral is rather messy. Doing the inside integral we get 30(2 — Z) + (2 — 2)?
Integrating that in the next stage gives §(4 — x2)3/2 + 435(4 — 352)”2 + 8(4 — $2)1/2. Finally,
evaluating the dx integral, we get 207?. Problem 3 (a) For f(x, y) = x2 + yez, ﬁnd the gradient ﬁeld €15.
ANSWER: grad( f) = V f = 125+ fyj+ f2]? = 2967+ 67+ yezl; (b) Let EQC, y, ,2) be the vector ﬁeld 2xT+ ezf+ yea/g. —> (i) Find div(F).
ANSWER: diU(F) 2 AIM?!— Nyj—l— P2]? where A] = 2x, N = 62, and P = yez. Hence
dz’v( F) = 2T+ 07+ gen; (11) Find curl(E).
ANSWER: We could conipute this directly. But if we compare F to the answer to part (a), we see that F =gmd(f). Since curl(gmd(f)) is zero for any f, the answer
is 6 2 07+ 0j+ 0k. Problem 4 Find all local niaxirna, local niinirna, and saddle points, for f (x, y) = 6x2 — 2x3 + 3y2 + 6mg.
Be sure to give both the points at which f takes on the values and the values it takes on there.
ANSWER: There is no boundary speciﬁed and f as a polynomial is continuous everywhere, so
the only candidates are points where the partial derivatives are both zero. Computing, f9; =
1235 — 6x2 + 6y and fy 2 63; + 636. If fy = 0, 6y + 6x = 0, i.e. y 2 —3c. Substituting that in
fw =Owe have 12m—6x2—6x20, so 6m—6m2 26$(1—m) =0. Hencem=00r$= 1. If
x = O, y = —x = 0, while if x = 1, y = —x = —1. Thus the two candidate points are (0, O) and
(1, —1). To check each of these as to whether it gives a saddle point or a local maximum or minimum,
we use D = fmfyy — 3y. fm = 12 — 12x, fyy = 6, and fxy = 6. At (0,0), D = 36 > 0 so
there is either a local maximum or a local minimum here. Since f$(0, 0) = 12 > 0, it is a local
minimum. The value of f at this point is f(0, 0) = 0. At (1, —1), D = —36 < 0, so there is a
saddle point. If we look at points along the x—axis, where y = 0, f(x, 0) = 6x2 — 2x3. (For very
large positive values of .76 this becomes arbitrarily large and negative, while for large negative
values of x it becomes large and positive. Hence there is no global maximum or minimum for
the whole plane.) Problem 5
Evaluate the line integral / (1 — m)d8 where C is a portion of the circle of radius 2 and center
0 (0,0), traversed from (2,0) to (0,2).
ANSWER: The curve can be parametrized as m = 2cost, y = 2sint, 0 S t S The function 1 — m becomes 1 — 2 cost. y’ = 2 cost and m’ = —2 sint. We can evaluate the integral as
/§(1 — 2cost)\/4sin2t + 4costht = 2/§(1 — 2cost) dt = 2 [t — 2sint]0g 2 7T — 4.
0 0
Problem 6
What are the largest and smallest values that f (m, y) = my takes on, for (m, y) on the ellipse
$2 y2
— — = 1?
8 + 2 At what points does f achieve those values?
ANSWER: We can view this as a constrained maximum / minimum (Lagrange multiplier) prob— lem with the objective function f(m, y) = my and constraint g(m, y) = “38—2 + g — 1 = 0. Then $1” = yT+ mﬁ and $9 = fi—l— yj To ﬁnd where these are parallel we solve 6f 2 A($g)
for some value A. We need to ﬁnd m and y (and A perhaps as a tool for ﬁnding m and y) that satisfyy=A(§),m=Ay,and%+%—1=O. From the ﬁrst equation we get % = 2, while the second yields % = (We divided by m and by
A: If A = 0 then both m and y are 0, which does not ﬁt g(m, y) = 0, so we can assume A 7A 0. If
m = 0 and y = A with A 7E 0, then y = 0, so again we have the point (0,0) which is not on
the ellipse.) Setting these equal we have 2 = i or A2 = 4. Hence A = :2. If A = 2, m = Ay gives m 2 2y. Using g(m,y) = 0 we have % + g = 1, y2 = 1, so y = ::1. If
y = 1, m 2 2y 2 2, so we have the point (2,1). If y = —1 we get the point (—2, —1). If A = —2, so m 2 —2y, we get in the same way the points (—2, 1) and (2, —1). Now we have four points to consider, ( 2, 1). We evaluate the function f(m, y) = my at each
of these. At (2,1) we get f(m,y) = 2. A (2,—1) we get f(m,y) = —2. At (—2,1) we get
f(m,y) = —2. At (—2, —1) we get f(m,y) = 2. Hence the maximum value, f(m,y) = 2, occurs
at the points (2, 1) and (—2, —1), and the minimum value, f(m, y) = —2, at (2, —1) and (—2, 1). Problem 7
A thin plate covers the region in the plane bounded by the m—aXis7 the line x = 17 and the curve 3/ = ﬂ, Where y 2 0.
The density of this plate is given by the function 6(36, y) = x + y. (a) Use an integral to evaluate the mass of this plate. ANSWER: The plate is shown at the right. We compute the mass by
integrating the density function through the region, 1 1 1 1 y4 13
M=// dd=/— ___3d:_
0 y2($+y)$y 0<2+y 2 y 96 20 (b) Find the moment My; of the plate about the m—axis.
ANSWER: M 1 1 d d 3
m—/0 AZWHW) x y—l—O (c) Find the moment Ag of the plate about the y—aXis.
ANSWER: A] 1 1 d d 19
311—1142347544» 95 y—E (d) Find the coordinates (E, y) of the center of mass of this plate.
ANSWER:  — _ _ m
The .76 coordinate x — M — 273. The y coordinate y = = %. Problem 8 Find the volume of the region that is above the cone
9b = g and inside the sphere p = 1.
At the right is a ﬁgure cut away so that you can see inside the sphere. ANSWER: We set up the integral in spherical coordinates. The radius p goes from 0 at the
center out to 1 on the sphere. The horizontal direction 9 goes all the way around from 0 to 27?.
And the vertical angle p goes from 0 at the top, along the z—aXis, down to Hence the volume
integral is 271’ 1 g 2 . 71—
/ ps1ngz§dgz5dpd6=—.
0 0 0 3 ...
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 Fall '08
 DICKEY
 Multivariable Calculus, dy dx, dy dm

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