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Unformatted text preview: MA 261 EXAM 2 Spring 2003 Name ___.___._..‘.._.__—————————'—‘ Student ID Recitation Instructm _____._————— Recitation Time _ Directions 1. Write your name, student ID number, recitation instructor’s name and recitation time
in the spaces provided above, and also ﬁll in the information on the answer sheet. 2. Circle the letter of your answer for each question in the test papers and also on the
answer sheet. 3. The exam has 12 problems. Problems 1—4 are worth 9 points each. All others are
worth 8 points each. 4. No books, notes or caICulators may be used in this exam. MA 261 EXAM 2 SPRING 2003 1. Given m2+y2+sin(a:y2) = 1. Find 3—: at. (0,1).
A. 0
1
B. E
C. 2
1
D ‘5
E. —2 2. Find the directional derivative of ﬂay) = tan(x + 23;) at the point (0, g) in
direction of (—3, 4). D. 16 E. —20 3. Critical points of f(:c,y) = 1:3 + y3 — 633; are . a saddle point and a local maximum . a saddle point and a local minimum A
B
C a local maximum and a local minimum
D. two saddle points and a local maximum
E . two saddle points and a local minimum 4. Use Lagrange multipliers to ﬁnd the point (3:, y, 2) at which 3:2 + y2 + 22 is minimal subject. to :3 + 23; +32: = 1.
1 1 2
A. _ _ _
(7’ 14’ 7) 3, 2,1
14 7 14 5. What is the double integral for the volume of the solid in the ﬁrst octant bounded by
the surfaces z2 = 3:2 + 2y”, 2: + y = 1. 6. Evaluate ff
R “92:2
1 + y2 (M, R: [—3,3] x [0,1]. 1 1 —:n
A. f f (x2+2y2)dydx
o D 1 1—2:
B. f f (x2+2y2)2dydz
o a
1 1—};
C. f f V322 + 2y2dyda:
0 o
1 1+5:
1). f f \/ 3:2 + 2y2dydm
0 o
1 1—2:
E. f / Mac? + 2y2dyda:
o a 18 91112 apow? 181112 1 1
7. Evaluate f f V 9:3 + 1 dxdy.
0 ﬂ A.%@§—n
B.§@%—n C. cannot be evaluated
0.; E.%@%—n 8. Evaluate ff my dxdy where R is the region in the ﬁrst quadrant that lies between
R the circles $2+y2=1 and $2+y2=9. 10 1011' POP”?
to
D 9. A lamina occupies the region in the ﬁrst quadrant bounded by y = :62 and y = 1. If
density p(:c,y) = my, the .1: coordinate of center of gravity equals 13.;
0.3
0.; 10. Find the area of the part of the surface z = y2 — 23: that lies above the triangle with
vertices (0,0), (1,1), and (%,1). 1 A. a (27 — 5J5)
1 B. E (27 — 5J5) 1
02 1
D. ﬁ(6\/6—2 2 E. ili (6J6 — 2J5) 11. Find the volume of the solid bounded by the surface 3; = 3:2, 2: = 0, and y + z = 1. A. calm (ﬁlm 12. Evaluate ff] de where E lies between the spheres $2 + y2 + 2:2 = 1 and
E
2:2 + y2 + Z2 = 4 in the ﬁrst octant. 7_1r
12 ...
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 Spring '08
 Stefanov

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