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UNIVERSITY OF MALTA FACULTY OF ENGINEERING B.Eng. (Hons.) Year III January 2015 Examination Session MAT3815 Mathematics for Engineers III 24th January 2015 9.15 a.m. - 11.15 a.m. Answer ONLY three questions. Only the booklets of mathematical formulae and calculators provided by the Faculty of Science can be used. Useful vector identities are included at the end of this paper. 1. (a) The volume of a rectangular tank open at the top is 4 cubic meters. If the base of the tank measures x by y meters, show that the tank’s surface area A in square meters is xy + 8 y + 8 x . Find the dimensions of the tank for A to be a minimum. 17 marks (b) If A, B and C are the angles of a plane triangle, find by using the method of Lagrange’s multipliers, the maximum value of the function f (A,B,C ) = sin A sin B sin C . 18 marks 2. (a) State Gauss theorem. An electrostatic vector field E is defined throughout a region V bounded by the closed surface S . An associated scalar potential function φ is given by E = -gradφ, where it is assumed that φ vanishes on the boundary S . The total energy W of the field is given by W = 1 2 Z V ρφ dV, where the charge density ρ is determined by the equation divE =4πρ. Deduce that W = 1 8π Z V E 2 dV, where E = |E|. (Hint: In this part of the question you may need to use one or more of the vector identities.) 18 marks Page 1 of 3

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UNIVERSITY OF MALTAFACULTY OF ENGINEERING

B.Eng. (Hons.) Year IIIJanuary 2015 Examination Session

MAT3815 Mathematics for Engineers III 24th January 20159.15 a.m. - 11.15 a.m.

Answer ONLY three questions.Only the booklets of mathematical formulae and calculators provided by theFaculty of Science can be used.

Useful vector identities are included at the end of this paper.

1. (a) The volume of a rectangular tank open at the top is 4 cubic meters.If the base of the tank measures x by y meters, show that the tank’ssurface area A in square meters is xy+ 8

y+ 8

x. Find the dimensions of

the tank for A to be a minimum.

17 marks

(b) If A, B and C are the angles of a plane triangle, find by using themethod of Lagrange’s multipliers, the maximum value of the functionf(A,B,C) = sinA sinB sinC.

18 marks

2. (a) State Gauss theorem.An electrostatic vector field E is defined throughout a region V boundedby the closed surface S. An associated scalar potential function φ isgiven by E = −gradφ, where it is assumed that φ vanishes on theboundary S. The total energy W of the field is given by

W =1

2

∫V

ρφ dV,

where the charge density ρ is determined by the equation divE = 4πρ.Deduce that

W =1

∫V

E2 dV,

where E = |E|.(Hint: In this part of the question you may need to use one or moreof the vector identities.)

18 marks

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(b) Evaluate ∫C

(sinx+ y2) dx+ (x− e−y) dy,

where C is the boundary of the semicircular region x2+y2 ≤ 4, y ≥ 0,taken in the anticlockwise sense.

17 marks

3. (a) Using cylindrical polar coordinates (ρ, φ, z) or otherwise, evaluate thetriple integral

3

16π

∫∫∫Q

z dxdydz,

where Q is the solid bounded by the paraboloid z = x2 + y2, thecylinder x2 + y2 = 4 and the xy-plane.

17 marks

(b) State Stokes’ Theorem.If φ is a scalar field and B is a vector field defined on an open surfaceS bounded by a closed curve C, show that∫

S

φ curlB.dS =

∫C

φB.dr−∫S

(gradφ ∧B).dS.

Hence if B = gradψ show that∫S

(∇φ ∧∇ψ).dS =

∫C

(φ∇ψ).dr = −∫C

(ψ∇φ).dr.

18 marks

4. (a) Find the work done by the force F = (yz + 2z)i + xzj + (xy + 2x)kin moving a particle from the point A(1, 0, 1) to the point B(0, 1, 1)along a path consisting of two straight line segments AC and CB,where C is the point (0, 0, 1).

17 marks

(b) Evaluate∫SA.n dS, where A = 4xi − 2y2j + z2x2k over the curved

surface S of the open cylinder x2 + y2 = 4, 0 ≤ z ≤ 3, where n is theoutward normal to S.

18 marks

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Useful Vector Identities

In the following identities F, G, and H are vector fields and φ is a scalarfield.

1. div(φF) = φdivF + F.gradφ

2. div(F ∧G) = G.curlF− F.curlG

3. div(curlF) = 0

4. curl(gradφ) = 0

5. curl(φF) = φcurlF + (gradφ) ∧ F

6. curlcurlF = graddivF−∇2F

7. grad(F.G) = F ∧ curlG + G ∧ curlF + (F.∇)G + (G.∇)F

8. curl(F ∧G) = FdivG−GdivF + (G.∇)F− (F.∇)G

9. F.(G ∧H) = H.(F ∧G) = G.(H ∧ F)

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