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Calculus III - Final Exam Solutions: Vector Calculus and Line Integrals, Exams of Advanced Calculus

University of AllahabadAdvanced Calculus

Solutions to the final exam of ma 227 (calculus-iii) focusing on vector calculus and line integrals. It includes problems on finding work done by force fields, determining conservative vector fields, using green's theorem, and evaluating surface integrals.

Typology: Exams

2012/2013

Uploaded on 03/16/2013

lahar
lahar 🇮🇳

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MA 227 (Calculus-III) Final exam
Show your work. Each problem is 10 points. Tue, Dec 14, 2004
Solve 10 problems for full credit. You can do all 11 problems for extra credit.
1. Find the work done by the force field F=yi−2xjaround the upper semicircle
{x2+y2= 9, y ≥0}, oriented in the counterclockwise direction.
Answer: ZC
F·dr=Zπ
0
(−9 sin2θ−18 cos2θ)dθ =−27π
2
pf3
pf4
pf5
pf8
pf9
pfa

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Download Calculus III - Final Exam Solutions: Vector Calculus and Line Integrals and more Exams Advanced Calculus in PDF only on Docsity!

MA 227 (Calculus-III)Show your work. Each problem is 10 points. (^) Tue, Dec 14, 2004Final exam Solve 10 problems for full credit. You can do all 11 problems for extra credit.

  1. Find the work done by the force field F = yi − 2 xj around the upper semicircle {x^2 + y^2 = 9, y ≥ 0 }, oriented in the counterclockwise direction. Answer: (^) ∫ C^ F^ ·^ dr^ =

∫ (^) π 0 (−9 sin

(^2) θ − 18 cos (^2) θ) dθ = −^272 π

  1. (a) Determine whether or not(b) If it is, find a function f such that F = (2 Fx =− ∇y)fi .− xj is a conservative vector field. (c) Compute the line integral ∫ C F · dr, where C is the line segment from (1, 2) to (2, 1).

(a) SinceAnswers: ∂P ∂y =^ −^1 and^

∂Q

∂x =^ −^1 (they are equal) the field is conservative. (b) f = x^2 − xy. (c) f (2, 1) − f (1, 2) = 3.

  1. Use Green’s theorem to find the area bounded by the curve r(t) = (2 − 3 cos t)i + (5 + 2 sin t)j, 0 ≤ t ≤ 2 π

the area must be negated:^ Answer: note that the orientation is clockwise! Therefore the standard formulas for

A = −

∫ (^2) π 0 (2^ −^ 3 cos^ t)(2 cos^ t)^ dt^ = 6π

  1. Determine whether or not the vector field F = (2xy − z^2 )i + (x^2 − 2 yz)j − (y^2 + 2xz)k is conservative. If it is, find a function f such that F = ∇f. Answer: since curl F = 0, the field is conservative. Then f = x^2 y − xz^2 − y^2 z.
  1. Use Stokes’ theorem to evaluate ∫ C F · dr, where F = (y + ez^ )i + (3x − z sin y)j − (z + ln(x^2 + 1))k and C is a triangle with vertices (0, 0 , 0), (1, 0 , 0), and (0, 1 , 0) traversed in this order. Answer: first of all, we need to compute curl F = 2k. Then ∫ C^ F^ ·^ dr^ =

D^ curl^ F^ ·^ n^ dA^ =

0

∫ (^1) −x 0 2 dy dx^ = 1

  1. Use the Divergence Theorem to evaluate ∫∫ S F · dS, where F = (x^3 − yz^5 )i + (ez^ − 3 x^2 y)j − (2z + xy−^2 )k and S is the sphere x^2 + y^2 + z^2 = 0.25 oriented with outward normal vectors. Answer: ∫ ∫ S^ F^ ·^ dS^ =

E^ div^ F^ dV^ =

E^ (−2)^ dV^ =^ −2 Vol(E) =^ −π/^3

  1. Evaluate the surface integral ∫∫ S F dS, where F = xi + yj + (5 − 3 z)k and has upward orientation. S is the part of the paraboloid z = 1 − x^2 − y^2 lying above the plane z = 0, and S

Answer: ∫ ∫ S^ F^ dS^ =

D

( −P ∂g ∂x −^ Q

∂g ∂y +^ R

) dA

=

D^ (5x

(^2) + 5y (^2) + 2) dA =^ ∫^2 π 0

0 (5r

(^2) + 2) r dr dθ =^92 π

11.segments: one starts at ( Evaluate the line integral− 1 , − (^) 1) and terminates at (3∫ C (x + y) ds, where C (^) , is a curve consisting of two line0), and the other starts at (3, 0) and terminates at (4, 4). Answer: parametric equations of these line segments are r 1 (t) = 〈 4 t − 1 , t − 1 〉 and r 2 (t) = 〈t + 3, 4 t〉 (for both, 0 < t < 1). Note that |r′ 1 | = |r′ 2 | = √17. Thus ∫ C 1 (x^ +^ y)^ ds^ +

C 2 (x^ +^ y)^ ds^ =

0

√17(5t − 2) dt +^ ∫^1 0

√17(5t + 3) dt = 6√ 17