261E2-F02 - MA 261 — Test 2 Name TA: Instructions 1. Each...

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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 mark-sense 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 3-1 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 div-d9 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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This note was uploaded on 04/28/2008 for the course MA 261 taught by Professor Stefanov during the Fall '08 term at Purdue University-West Lafayette.

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261E2-F02 - MA 261 — Test 2 Name TA: Instructions 1. Each...

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