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Unformatted text preview: MA 261 — Test 2 Name TA: Instructions 1. Each problem is worth 6 points except the last problem is worth 10 points. 2. Circle your choice of the correct answer and blacken the corresponding circle on the
mark~scnse sheet. 3. Calculators or books are not allowed. 4. Both the test booklet and the marksense sheet are to be given to the TA at the end
of the examination. Test2, MA 261 1. The directional derivative of f(:n, y) : 2:38—23" in the direction of greatest increaSe of f
at x z 11 y = 0 is —' A. 3:“ B. 33—23
C. 3 D. J5
E. x/ﬁ 2. Find the minimum value of f($, y) = 2:1: + 9 subject to the constraint 1‘2 + y? = 1. A. —1
B. —\/5
C. —2
I). 0
E. —»/3 3. Which of the following points corresponds to a local maximum of
f(.'E, y) = 61:192 — 2x3 — 319“. A. (0, 1)
B. (1, —2]
C. (1.1)
D. (—1,1)
E. (1, 0) 4. Evaluate :cgdA, where R is the rectangle 0 g :1: g 3, 0 g y _<_ 2.
R A 18
B. 9
C. 8
D 19
18
E. —
3 5. Find the volume of the solid in the first octant bounded by z = 4 — y2 and :1: = 1. A. 9
3
20
B. _
3
18
C. —
3
D. 1,—5
.3
22
E. _
3 1 2 6. The integral / [ ydzcdy is the volume of which region.
0 0 % 7. The area of the triangie with vertices (0,2), (0,1), and (1,0) is given by 2 A. f / dandy
U ‘1—y
2 2.1:
B. f / dydm
0 a:
1 2z+1
C. dydx
0 a:
1 2—ém
D. dydzr:
[I 1—3
1 2—2:
E. dydm
0 1—3: 8. After an interchange of the order of integration the integral [26 [1 its, mdyda: equals 3 6
A. [I f(1=,y)dxdy 2:: B. 3 [yﬂmmwxdy
0— [13 1f(w,y)dxdy D. [26f(z,y)dscdy E. f: [I f($,y)d:rdy 1 W
9. Evaluate [ / cos(x2+y2)dydw. —10 P" E]. B
C.
D 31
25K112 «c031 7r
—'1
25m 7rsin1 w
—c031
2 10. Find the surface area, of the part of the surface 2 z :1: + y2 that lies above the triangle with vertices (0,0), (0, 2), and (2, 2). 1 .1 my
11. Evaluate f f / (2x + 8yz)dzdydac.
o o o A. ——13.‘/§
3
26
B. —
\/§
C {Hf—21
D. 1.3
3
17\/§
E. T
10
A a
3
B. §
11
C. E
9
D I?
E. E 12. Find fix/:62 + y? + z2 dV where H is the solid hemisphere with center the
H origin, radius 1, that lies above the Iy—Iilane. A. if
7
B 2w
Tl."
0 n
D. 3r—
25
E. 5—“
9 § Ssecﬂ
13. The region of integration of the iterated integral f j r divd9 is
e e A. a rectangle B. inside of part of a rose curve
C. inside of part of a cardieid
D. a triangle E. a circular sector 14. Which integral gives the volume of the solid in the ﬁrst octant bounded by the surfaces
$2+z2:9,y=21r,y=0,z=0: 3 5
A. f / V9 — yzdydx
0 e
3 2:
B. f f V 9 r mZdyda:
e 0 6 23 I
C. / / (5:2 + z2)dyd:rr
o o 6 s
D. / j V 1 — Izdyda:
e e 3 2::
E. f f (1:2 + z2)dyda:
e e 15. Fill in the quantities a and b that convert the triple integral from rectangular coordi
nates to spherical coordinates: ‘2 VII—at! \/1:2+y2 'rr % 2c3cgo
/ f / zdzdydsc : / f / bdpdgodﬂ.
—'2 0  0  0 a. CI
A a = 0, b : p3sinzpcos<p B. a=0, b=p2cos<p ar
C. a=—, b=p33ingo 4 7r 3 .
D. a" Z, b=p cosgosmrp
E. (1:0, b=pCOStp If$=u—v,y=u+2v,then/f(y—:r)dA=
R 1 2—21;
A. / / devdu
u o
1 2—21;
B. / / 3Ud'udu
o 0
1 2
C. f / devdu
0 o
1 2
D. f / devdu
0 2—2u
1 (23“)
E. / [ Bydvdu
o o ...
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 Fall '08
 Stefanov
 TA, B., E., Multiple integral

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