PROBLEM: Evaluate d A, where is the boundary of the cube defined by -1 x 1, -1 1, and 0 z 2.REFERENCE: http://math.bard.edu/~mbelk/math601/GaussExamples.pdfCOPIED BY: NOVILLA, ASHNERSOLUTION:div ( = + ( + ( = 3y2 + xThe divergence theorem gives: d A = = = 2 = 8

PROBLEM: Evaluate both sides of the divergence theorem for the field D = 2xy ax + x2 ay C/m2 and the rectangular parallelepiped formed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.REFERENCE: http://math.bard.edu/~mbelk/math601/GaussExamples.pdfCOPIED BY: NOVILLA, ASHNER P.SOLUTION: +

= However, = 0 and =, which leaves only = = = Since (2xy) + (x2) = 2yThe volume integral becomes = =

PROBLEM: Suppose C is the curve obtained by intersecting the plane z = x and the cylinder x2 + y2 = 1 oriented counter clockwise when viewed from above. Let S be the inside of this ellipse, oriented with the upward-pointing normal. If F = x ax + z ay + 2y az, verify Stokes' theorem by computing both and REFERENCE: COPIED BY: NOVILLA, ASHNERSOLUTION:The line integral, The ellipse is a graph (using z = x) over the unit circle in the xy-plane. Thus we can parameterize it as r(t) = (cos(t); sin(t); cos(t)) for 0. Since F = (x, z, 2y), we getF(r(t)) = (cos(t), cos(t), 2 sin(t)) dr = ( - sin(t), cos(t), - sin(t)) dtand soF(r(t)) dr = sin(t) cos(t) + cos2(t) - 2 sin2(t)dt = sin(t) cos(t) + 1 3 sin2 (t) dtThus = = - The flux integral Again the elliptical disk is a graph (using z = x) over the unit disk in the xy-plane. Thus we can parameterize it as r(x, y) = (x, y, x) for x2 + y2 Since F = (x, z, 2y), we get curl F = (1, 0, 0) and rx X ry = (-1, 0, 1). Thus = - = - PROBLEM: Suppose S is a \light-bulb-shaped region" as follows. Imag-ine a light-bulb cut o at the base so that its boundary is the unit circle x2+ y2= 1, oriented with the outward-pointing normal. (You can use either an old-fashioned light-bulb or a compact Fluorescent if you're feeling green.) Suppose F= Compute the flux integral

REFERENCE: COPIED BY: NOVILLA, ASHNERSOLUTION:The point of this problem is to find use Stokes' theorem to avoid computing the flux integral over S (whatever confusing surface that could be) and instead compute the line integral over the unit circle C in the xy-plane. We use the parameterization r(t) = (cos(t), sin(t), 0), soF(r(t)) = (cos(t), sin(t) + 1, 0)dr = h sin(t), cos(t), 0) dtF(r(t)) dr = cos(t) dt.Thus = =

LaplacianPROBLEM: Find the Laplace transform of the following functions.(a) (b) (c) (d) REFERENCE: Cazelais, G. (2006).Laplace Transform. [ONLINE] Available at: http://cazelais.disted.camosun.bc.ca/175/laplace-book.pdf. [Last Accessed September 10,2012].COPIED BY: NOVILLA, ASHNER GERALD P.Solution:(a) (b) (c) (d)

PROBLEM: The differential equation for the current i(t) in an LR circuit is L di/dt+Ri=V(t). Find the current in an LR if the initial current is i(0)=0A given that L=2H,R=4 and V(t)=5e^(-t) volts with time t measured in seconds.REFERENCE: Cazelais, G. (2006). Laplace Transform. [ONLINE] Available at: http://cazelais.disted.camosun.bc.ca/175/laplace-book.pdf. [Last Accessed September 10,2012].COPIED BY: NOVILLA, ASHNER GERALD P.SOLUTION:The differential equation of the LR circuit is 2i^' (t)+4i(t)=5e^(-t)With initial condition i(0)=0. Take the Laplace transform on both side of the differential equation to get2sI(s)+4I(s)=5/(s+1)Solving for I(s) gives I(s)=5/((2s+4)(s+1))Using partial fractions, we obtainI(s)=5/(2(s+1))-5/(2(s+2))Taking the inverse Laplace transform gives the answeri(t)=5/2 e^(-t)-5/2 e^(-2t)

Answer: The current is i(t)=5/2 e^(-t)-5/2 e^(-2t)

PROBLEM: Find the gradient of the following scalar field U =.REFERENCE: Retrieved from http://elcondensador.net/media/ElectromagnetismoI/chapter_03.pdf (2012)SOLVED BY: NOVILLA, ASHNER GERALD P.SOLUTION:

Answer:

a) The temperature at point A is 10 C, the temperature at point B is 20 C, and point A is 100 km from point B. What is the temperature gradient?

dT = 20 C - 10 C = 10 C, dx = 100 km; dT/dx = 10 C / 100 km = 0.1 C/km

b) The pressure at point A is 1020 mb, the pressure at point B is 995 mb, and point A is 400 km from point B. What is the pressure gradient?

dP = 1020 mb - 995 mb = 25 mb, dx= 400 km, dP/dx = 25 mb / 400 km = 0.0625 mb/km, or 6.25 mb per 100 km