Exam 3 Solved Questions and Cheat Sheet - Calculus 3 for Engineers | APPM 2350, Exams of Mathematics

Download Exam 3 Solved Questions and Cheat Sheet - Calculus 3 for Engineers | APPM 2350 and more Exams Mathematics in PDF only on Docsity! APPM 2350 EXAM 3 FALL 2009 INSTRUCTIONS: Books, notes, crib sheets, and electronic devices are not permitted. Write your (1) name, (2) instructor’s name, and (3) recitation number on the front of your bluebook. Work all problems. Show and explain your work clearly. Note that a correct answer with incorrect or no supporting work may receive no credit, while an incorrect answer with relevant work may receive partial credit. 1. (25 points) Determine the volume of material removed from the solid sphere x2 +y2 + z2 ≤ 9 by the cylinder r = 3 cos θ. 2. (25 points) Consider the region in the xy-plane bounded by the closed curve 4x2 + 4xy + 2y2 + 4y = 5. (It is actually an ellipse.) You need to find the enclosed area A. The substitution u = 2x+ y and v = y+ 2 will simplify your calculation. (a) Find x and y in terms of u and v using the given substitution. Be sure to check this because the rest of the problem depends on this result! (b) Transform the original region Rxy into its corresponding region Ruv in the uv-plane. Make a clear sketch, of the new region of integration Ruv in the uv-plane. Be sure to label all axes, boundaries, intersection points, etc. on your sketch. (c) Rewrite the integral for A over the region Ruv in the uv-plane in terms of u and v. (d) Evaluate A in terms of u and v. 3. (25 points) The integral V = ∫ 2π θ=0 ∫ 1 z=0 ∫ √2 r=z r dr dz dθ calculates the volume of an object. (a) Make a clear sketch of the cross-section of the object in a rz-plane (this is a constant θ plane in cylindrical coordinates) clearly labeling the bounding surfaces of the region of integration. (If you have trouble with this, you may “buy” a sketch of the shape of the region in the rz-plane for 5 points. This sketch will only show the shape of the region, so you will still need to supply the remaining details. The offer to buy this sketch ends at 6:15 PM!) (b) Express V in cylindrical coordinates using the order dz dr dθ. (c) Express V in spherical coordinates using the order dρ dφ dθ. (Hint: if needed, you should refer to particular angles as φ = arctan(a/b). For example, φ = arctan( √ 2/ √ 3).) (d) Express V in spherical coordinates using the order dφ dρ dθ. (e) Evaluate one of the integrals above to determine the value of V . 4. (25 points) Consider the path (section 1) starting at the origin straight along the x-axis to the point (3, 0), then (section 2) through the first quadrant along the curve x2 9 + y2 4 = 1 from the point (3, 0) to the point (0, 2), and the vector function given by F = −4y i + 2x j. (a) Sketch the entire path and give a parametrization for each section of the path. (b) Calculate the flow along each section of the path C. Be sure to clearly state what the flow is for each section. (c) Calculate the flux along each section of the path C. Be sure to clearly state what the flux is for each section. OVER Projections and distances projAB = ( A ·B A ·A ) A d = |−→PS × v| |v| d = ∣∣∣∣−→PS · n|n| ∣∣∣∣ Arc length, frenet formulas, and tangential and normal acceleration components ds = |v| dt T = dr ds = v |v| N = dT/ds |dT/ds| = dT/dt |dT/dt| B = T×N dT ds = κN dB ds = −τN κ = ∣∣∣∣dTds ∣∣∣∣ = |v × a||v|3 = |f ′′(x)| | 1 + (f ′(x))2 |3/2 = |ẋÿ − ẏẍ| |ẋ2 + ẏ2|3/2 τ = −dB ds ·N a = aNN + aTT aT = d|v| dt aN = κ|v|2 = √ |a|2 − a2T Directional derivative, discriminant, and Lagrange multipliers df ds = (∇f) · u fxxfyy − (fxy)2 ∇f = λ∇g, g = 0 Polar coordinates x = r cos θ y = r sin θ r2 = x2 + y2 dA = dx dy = r dr dθ Cylindrical and spherical coordinates Cylindrical to Rectangular Spherical to Cylindrical Spherical to Rectangular x = r cos θ r = ρ sinφ x = ρ sinφ cos θ y = r sin θ z = ρ cosφ y = ρ sinφ sin θ z = z θ = θ z = ρ cosφ dV = dx dy dz = r dr dθ dz = ρ2 sinφdρ dφ dθ Substitutions in multiple integrals∫ ∫ R f(x, y) dx dy = ∫ ∫ G f (x(u, v), y(u, v)) |J(u, v)| du dv where J(u, v) = ∂x ∂u ∂y ∂v − ∂y ∂u ∂x ∂v Mass, moments, and center of mass Mass M = ∫ ∫ R δ dA Moments Mx = ∫ ∫ R y δ dA My = ∫ ∫ R x δ dA Center of mass x̄ = My/M ȳ = Mx/M Flow and flux Flow = ∫ C F ·T ds = ∫ C F ·V dt = ∫ C F · dr = ∫ C M dx+N dy Flux = ∫ C F · n ds = ∫ C M dy −N dx om ‘ Ge dee dug + Uh thy eh Core As elco b) £= (249)" 4 (qaey 24 : = want= 4 Se kw uw plan othe Pegram of Larearasti- As @ ards wrth, Cats 4 | ~ urd vast al UF e+ y KF 7 = te- vs | 2 MF gal ya V-2z © Xe YT 4 4 = 4 4) © Tou) > | = | = 4 Ny 4% o w R eb’ mam 3 on = drte - x |(Z)de A L r ¥ v (x e*0 yo eg \ “ £ (2) = W ue 3a | e } wo v= | | [ vaedede | | [rasadeate 6:0 fo Bo oo +41 &e 2 ™ r° Sk He 4 wl | } |e ag hfde * j j | extataeete 1 we FH sg o eee Seo oon (64) J ws \ i oat def dy ld * i r [soe (GO SO EME, So ost feces'(4) 2) fom cnet ve f I [car dea “ Bo ret Le - | c0-2)aade | da $e 679 250 ero oe —  x Bi © (2) os tee | re)= bOI (os ¢s)) LW) Beat) © Lech Ihe vey = 34 & Yt ant) + let | | t ) | flow, 2 | eet = { Feyae = | “443 at (s:0 ~©) i 7 ve | = |-12-0k 2=C @ re | loog . | sovat « [Grace 4 Yecat) he | ® a i mA | . a wt + (eat at = 12] ((asce te ; to Co Ye = fact lat =Gn ru -@ 2] te tte = Qy a flue © | ets , \ aot « 23t7 \e( of -3)de 3 “ & SG eo i (he = [ince -Q to . a ~ ‘Res x ; flan = jh * \(a-esst’ * 2t0t} tet saxty) a i annette GeO tit y | © ¢ Ea® Was itt Wo Yo = [letescdeat + \Reotemt be to Mt, & A = | sutttdey * st = ({ g— tt to Ta ett