Math 2E elected Problems for the Final Aaron Chen pring 216 These are the problems out of the textbook that I listed as more theoretical. Here s also some study tips: 1) Make sure you know the definitions of all the different integrals and objects. Mainly, understand all the bold faced words and boxed equations in the sections we ve covered. 2) Make sure you understand your past homework problems, quizzes, and the practice final. 3) Understand the reasoning behind most of the extra theoretical problems. 4) A nice way to recap a lot of these things is to go through the Review sections in Chapter 15,16. I recommend reading the relevant concept check questions and doing the true false questions, too: Chapter 15: Pg. 161. Chapter 16: Pg. 1148. isclaimer: The following solutions are only very sketched out! ome aren t even solutions, but just ideas or hints for the solutions. 16.3 Fundamental Theorem of Line Integrals #29 This is basically that if a vector field is conservative, then F. #3 This integral results from taking C F dr where F(x, y, z) < y, x, xyz >. The curl of this is not zero, î ĵ ˆk F x y z (xz )î + ( yz)ĵ + (1 1)ˆk y x xyz where the first two components are not zero. #35 a) The partials are P y (x2 + y 2 ) + y(2y) (x 2 + y 2 ) 2 y2 x 2 (x 2 + y 2 ) 2, Q x (x2 + y 2 ) x(2x) (x 2 + y 2 ) 2 y2 x 2 (x 2 + y 2 ) 2 so we see they equal. b) This integral is not path independent if the curve circles the origin. We see that if C 1 is the upper half of the circle (of radius 1), then we parameterize C 1 as r 1 (t) < cos t, sin t > with t π, so the tangent is r 1 (t) < sin t, cos t > and C 1 F dr π F(r 1 (t)) r 1(t)dt 1 π + sin 2 t + cos 2 t cos 2 t + sin 2 dt π. t
But, for C 2, the lower half of the circle, we parameterize it as the same r 2 (t) < cos t, sin t > from t π, so the tangent is r 2 (t) < sin t, cos t >. Then, C 2 F dr π F(r 2 (t)) r 2(t)dt π π ( 1) 3 sin 2 t cos 2 t cos 2 t + sin 2 dt t 1dt π. π ( sin 2 t cos 2 t)dt We see that the value of these two integrals is different, so the integral is NOT path independent. This does not violate anything because the components of F, and their partials too, are not even continuous at the origin (,). The domain of R 2 without the origin is not simply connected, and hence this non-simply-connectedness violates Theorem 6 (on Pg 179). 16.4 Green s Theorem #21 a) First we need to parameterize the line. It would be given by r(t) (1 t) < x 1, y 1 > +t < x 2, y 2 > < x 1 + t(x 2 x 1 ), y 1 + t(y 2 y 1 ) >, t 1. Then, dx x (t)dt (x 2 x 1 )dt and dy y (t)dt (y 2 y 1 )dt. Then, C xdy ydx 1 (x 1 + t(x 2 x 1 ))(y 2 y 1 )dt x 1 (y 2 y 1 ) + (x 2 x 1 )(y 2 y 1 ) 2 1 (y 1 + t(y 2 y 1 ))(x 2 x 1 )dt y 1 (x 2 x 1 ) (y 2 y 1 )(x 2 x 1 ) 2 x 1 (y 2 y 1 ) y 1 (x 2 x 1 ) x 1 y 2 x 1 y 1 y 1 x 2 + y 1 x 1 x 1 y 2 x 2 y 1. b) The boundary of the polygon is precisely the line segments connecting each (x k, y k ) to the next (x k+1, y k+1 ) vertex - call each line segment L k, with the last line segment L n connecting (x n, y n ) to (x 1, y 1 ). Recall as a consequence from Green s Theorem that Area(omain ) dxdy x 2 dy y 2 dx since with P y/2, Q x/2, then Q x P y 1. In particular, the boundary of is precisely the union of each line segment, in that so when we compute the integral, Area() L 1 + dxdy +... + L 2 L n 1 + L n n k1 L k x 2 dy y n 2 dx 1 xdy ydx 2 L k k1 2
where applying part (a) to each L k, where if we write this out, 1 n x k y k+1 x k+1 y k 2 k1 1 2 [(x 1y 2 x 2 y 1 ) + (x 2 y 3 x 3 y 2 ) +... + (x n 1 y n x n y n 1 ) + (x n y 1 x 1 y n )]. c) Is a computation - use the formula with just n 5 vertices. A 1 2 [( ) + (6 1) + (2 ) + ( ( 2)) + ( )] 1 [5 + 2 + 2] 9/2. 2 #29 < y, x> Here, F(x, y). Recall that from 16.3.35, we saw that its components P y and x 2 +y 2 x 2 +y 2 Q x satisfied Q x 2 +y 2 x P y. Moreover, these components are smooth everywhere except at the origin. This is half of what we need for path independence. The main idea is that we do not enclose the origin. For any simple closed path that does not enclose the origin, we can enclose that curve in some region of R 2 that is simply connected. This means that F is conservative on and since the curve lies in, we can use the Fundamental theorem of line integrals. ince F is conservative, it is the gradient of some f on, in that F f on. The curve is closed, so the final and initial point, call it P are the same. Then, F und.t hm C F dr f(p ) f(p ). #31 ee the practice final Problem 6. 16.5 Curl and ivergence #19 Firstly, since G always, check it: G < x sin y, cos y, z xy > sin y sin y + 1. This doesn t even satisfy that G so such a vector field G will not exist. #2 This time, doing the same thing, G < xyz, y 2 z, yz 2 > yz 2yz + 2yz again so such a vector field G again will not exist... #21 ee the practice final Problem 2(b). Irrotational means that the curl is zero. 3
#22 Incompressible means that the divergence is zero. But here, F x f(y, z) + y g(x, z) + z h(x, y) since each component is independent of the variable that it is being differentiated with respect to. #26 ee practice final Problem 5(b). Other problems like 23-29 are of this flavor too, but very long :( I m not going to do them. #3-32 ince the professor has had a few problems using the r vector, it may be nice to look at these again. For example, for 32, 32) if F r/r p, find F. Is there a value of p such that the divergence is? Here, first rewriting in Cartesian and following the definition of r, F(x, y, z) < x, y, z > (x 2 + y 2 + z 2 ) p/2. The divergence is then (careful with factors of 2 and 1/2 cancelling) so thus F (x2 + y 2 + z 2 ) p/2 px 2 (x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p from quotient rule on / x + (x2 + y 2 + z 2 ) p/2 py 2 (x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p from quotient rule on / y + (x2 + y 2 + z 2 ) p/2 pz 2 (x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p from quotient rule on / z 3 (x 2 + y 2 + z 2 ) p/2 p(x2 + y 2 + z 2 )(x 2 + y 2 + z 2 ) p/2 1 (x 2 + y 2 + z 2 ) p 3 (x 2 + y 2 + z 2 ) p/2 p(x2 + y 2 + z 2 ) p/2 (x 2 + y 2 + z 2 ) p Consequently, if p 3, this is zero. F 3 (x 2 + y 2 + z 2 ) p/2 p (x 2 + y 2 + z 2 ) p/2 3 p (x 2 + y 2 + z 2 ) p/2. 4
# 33 To begin the proof using Equation 13 (the corollary to Green s Theorem), which states F ˆnds F da we start with the one term in the identity that has a line integral. Namely, f g ˆnds Eqn.13 (f g)da but now, recall on Quiz 9 #1(c) that (ff) f F + f F. In particular, when F g, this gives us (f g) f g + f 2 g where 2 g g xx + g yy + g zz. Hence, we have that f g ˆnds Eqn.13 (f g)da in other words just writing it cleanly, f g ˆnds ( f g + f 2 g)da ( f g + f 2 g)da where now if we subtract the integral of f g, we have Green s first identity, f g ˆnds f gda f 2 gda. As mentioned in the review, this is kind of like a 2- integration by parts on f 2 gda. It s like u f, dv 2 g and the evaluation at endpoints of uv b becomes the line integral a over just the boundary, f g ˆnds. As you might guess, there is also a 3- integration by parts type of formula, too. #34 To begin this proof, with using Green s first identity (#33), apply it to both C f g and C g f and we will result with the Left Hand ide because the f gda terms subtract and cancel each other. Mainly first rearrange Green s 1st identity to read as f g ˆnds f gda + f 2 gda. Then, we have that for C f g and C g f, applying Green s 1st identity to each: f g ˆnds f gda + f 2 gda. g f ˆnds f gda g 2 fda. Adding the two, the middle integrals of f g cancel, and we are left with (f g g f) ˆnds (f 2 g g 2 f)da. 5
#35 For this, to apply Green s first identity, the idea is that C ngds C g nds. The book s just trying to use new notation - don t get confused by it! Mainly now, this is exactly in the form to use Green s first identity with f 1, the constant function 1. With this, we see that the f g integral is zero because f is constant, f (1). The other integral of f 2 g is also zero because g is harmonic, 2 g by definition. 1 g ˆnds #33 1 2 gda + (1) (g)da where the gradient of a constant is zero, and 2 g because g is harmonic, #36 +. With this proof, since we are looking at f 2, this is a hint that we should have f g because that way we get f g f f f 2. Mainly, this pops up in Green s first identity as the f gda term. Now, using Green s identity, we can conclude this integral is zero because:. Remember we have f g and f is harmonic so 2 f and the integral of f 2 g is really now f 2 f.. ince f on the boundary, the line integral C f f nds is zero because C is exactly the boundary of the domain. f 2 da f f da where now applying #33 1st identity of Green, and rearranging terms, f( f) ˆnds f 2 fda. o, just rewriting, we have f 2 da f( f) ˆnds f 2 fda. The first integral of f f is because f on the boundary of. The second integral of f 2 f is because f is harmonic, i.e. 2 f, so then f 2 da. 16.6 Parametric urfaces # 51 First, always stick with definitions! Here, Area() def 1 + f 2 x + f 2 y dxdy. x 2 +y 2 R 2 6
ince f x, f y 1, we see that 1 + fx 2 + fy 2 3 then. Hence, Area() x 2 +y 2 R 2 1 + f 2 x + f 2 y dxdy 3 dxdy 3 x 2 +y 2 R 2 where using the area interpretation of integrals, 3 πr 2. x 2 +y 2 R 2 da o, all we can conclude is that Area() is 3 πr 2. We can also note that 1 + fx 2 + fy 2 1 so then in the same way, Area() πr 2 too. #58 Hence, πr 2 Area() 3πR 2. I did this one in section. #6 (and 59 is similar, I ll skip 59) **I dont think you ll need to know hyperbolic sine (sinh) and hyperbolic cosine (cosh)...** a) Here, we notice that aside from the scalings in front of the x, y, z parameterizations, we can make the trig identity sin 2 t + cos 2 t 1 and also cosh 2 t sinh 2 t 1 work (I know hyperbolic trig functions are not that common, but they work similarly, though not exactly, like regular trig functions). To deal with the scalings, we just consider the quantities x/a, y/b, and z/c similar to what would be done in #59. Then, because we need cosh 2 sinh 2 1, we expect to take x 2 a 2 + y2 b 2 z2 c 2 cosh2 u sin 2 v + cosh 2 u cos 2 v sinh 2 u cosh 2 u(sin 2 v + cos 2 v) sinh 2 u cosh 2 v (1) sinh 2 u cosh 2 u sinh 2 u 1. o, we see that the components of the surface parameterization satisfy z 2 c 2 y2 b 2 x2 a 2 1 which is the general equation of a hyperboloid of one sheet. ince the components satisfy the hyperboloid of one sheet s equation, it must be (part of) a one-sheet-hyperboloid. b) It looks like a nuclear silo. c) The first thing to note is that if 3 z 3, then with c 3, we must have 1 sinh u 1 which means that sinh 1 ( 1) < u < sinh 1 (1). econdly, v 2π because this is what is essentially making x, y rotate around the graph (the nuclear silo). 7
Now, for the area formula, we need r u r v. Here, derivatives with cosh and sinh are a little different from their regular trig function counterparts, { r u < a sinh u cos v, b sinh u sin v, c cosh u > so then and hence, r v < a cosh u sin v, b cosh u cos v, > r u r v < bc cosh 2 u cos v, ac cosh 2 u sin v, ab sinh u cosh u(cos 2 v + sin 2 v) > r u r v Then, for the area for part (b), Area 2π sinh 1 (1) v < bc cosh 2 u cos v, ac cosh 2 u sin v, ab sinh u cosh u > u sinh 1 ( 1) b 2 c 2 cosh 4 u cos 2 v + a 2 c 2 cosh 4 u sin 2 v + a 2 b 2 sinh 2 u cosh 2 u. b 2 c 2 cosh 4 u cos 2 v + a 2 c 2 cosh 4 u sin 2 v + a 2 b 2 sinh 2 u cosh 2 ududv. We can simplify this a little, and also plug in for a 1, b 2, c 3 from part (b), 2π sinh 1 (1) v u sinh 1 ( 1) ome observations, see # 61,63 Notice that #63 is on the practice final, as Problem 7. cosh u 36 cosh 2 u cos 2 v + 9 cosh 2 u sin 2 v + 4 sinh 2 ududv For #61, first find the intersection of the paraboloid and sphere. I think it s just a circle. Then, the key is to get the right θ and in particular φ bounds by using some trig (inverse functions). The circle intersection should restrick the φ bound nicely between and whatever φ-value the circle of intersection is at. 16.7 urface Integrals # 37 When our surface is the graph of y h(x, z), then we first parameterize our surface as { r x < 1, h x, > r(x, z) < x, h(x, z), z > r z <, h z, 1 >. Then, r x r z < h x, 1, h z >. More importantly, this is the correct normal as the ĵ component (in the y-direction) is 1. This means it goes to the left in the y direction. Now, when we assume F < P, Q, R >, we have that F d F(x, h(x, z), z) (r x r z )dxdz x,z < P (x, h(x, z), z), Q(x, h(x, z), z), R(x, h(x, z), z) > < h x, 1, h z > dxdz x,z ( P h ) x Q + R h dxdz. z x,z 8
# 38 This is basically identical to #37. There s one tiny, but important difference though! The normal needs to point in the positive î direction (in the +x direction this time!). Our surface is the graph of x k(y, z) now. We first parameterize our surface as { r y < k y, 1, > r(y, z) < k(y, z), y, z > r z < k z,, 1 >. Then, r y r z < 1, k y, k z >. More importantly, this is the correct normal as the ĵ component (in the x-direction) is +1. This means it goes towards us in the +x direction. Now, when we assume F < P, Q, R >, we have that F d F(x, h(x, z), z) (r x r z )dydz y,z < P (k(y, z), y, z), Q(k(y, z), y, z), R(k(y, z), y, z) > < 1, k y, k z > dxdz y,z ( P Q k ) y R k dydz. z # 48 y,z This is exactly Problem 8(a) on the practice final. # 49 We want to show that F d regardless of the radius of the sphere. Here, we first write F using x, y, z variables as c < x, y, z > F(x, y, z) (x 2 + y 2 + z 2 ) 3/2. Let us use that F d F ˆn d. Here, recall from discussion that n sphere (x 2 + y 2 + z 2 ) < 2x, 2y, 2z > so then when we normalize it, ˆn <x,y,z>. Then, x 2 +y 2 +z 2 F ˆn c(x2 + y 2 + z 2 ) (x 2 + y 2 + z 2 ) 2 c R 2 because we are on the surface of the sphere (here, R is the radius of ). Then, c F d F ˆn d R 2 d c R 2 d c urface Area() R2 and using the surface area of a sphere, c R 2 4πR2 4πc. This is a constant value, and more importantly independent of the radius. This verifies the claim that the flux of F across the sphere is independent of the radius of. 9
16.Review #13 ( #14 was on an old quiz I think..) To show F is conservative, let s find f such that F f. Here, integrating the x-component, f(x, y) g(y) + (4x 3 y 2 2xy 3 )dx g(y) + x 4 y 2 x 2 y 3. Then, f y 2x 4 y 3x 2 y 2 + 4y 3 has to hold, it needs to equal the y component of F. which means that f y g y (y) + 2x 4 y 3x 2 y 2 needs to equal 2x 4 y 3x 2 y 2 + 4y 3 g y (y) 4y 3 g(y) y 4 + K where K is a constant. Hence for f(x, y) y 4 + x 4 y 2 x 2 y 3 + K, we have that F f so by the Fundamental theorem of line integrals, F dr f(final) f(initial). C Here, the final point is r(1) (1 +, 2 1) (1, 1) and the initial point is r() (, 1). o, F dr f(1, 1) f(, 1) (1 + 1 1 + K) (1 + + K). C (And we see, the constant doesn t affect the answer). #21 Apply Green s theorem! We see that Q g(y) so then Q/ x. imilarly, P f(x) so P/ y and hence by Green s this turns into an area integral of just, which is. #22 First, start with that #23 2 (fg) 2 2 2 (fg) + (fg) + x2 y2 z 2 (fg) 2 x (f xg + fg x ) + 2 y (f yg + fg y ) + 2 z (f zg + fg z ) [f xx g + f x g x + f x g x + fg xx ] + [f yy g + f y g y + f y g y + fg yy ] + [f zz g + f z g z + f z g z + fg zz ] [f xx + f yy + f zz ]g + 2[f x g x + f y g y + f z g z ] + f[g xx + g yy + g zz ] g 2 f + 2 f g + f 2 g. The idea here is that applying Green s, we get that f y dx f x dy ( f xx f yy )da and since f is harmonic, 2 f, da. 1 2 fda
# 29 Only because this is one of the special cases, a sphere (or cylinder) where we know d already. I ll do this in two ways, to hopefully clarify things. 1) tandard Way: We know what we parameterize the sphere as r(θ, φ) < R cos θ sin φ, R sin θ sin φ, R cos φ >. Then for the outward normal, we have to cross r φ and r θ, î ĵ ˆk r φ < R cos θ cos φ, R sin θ cos φ, R sin φ > r θ < R sin θ sin φ, R cos θ sin φ, > r φ r θ < R 2 cos θ sin 2 φ, R 2 sin θ sin 2 φ, R 2 cos φ sin φ(cos 2 θ + sin 2 θ) >. o reducing with trig identities, r φ r θ < R 2 cos θ sin 2 φ, R 2 sin θ sin 2 φ, R 2 cos φ sin φ >. (Indeed, this has outward orientation, all the components are positive). Then, since we integrate over the entire sphere, we have that after plugging in for x, y, z as functions of θ, φ, 2π π θ φ 2π π F d 2π π F(r(θ, φ)) (r φ r θ )dφdθ < R 2 cos θ sin φ cos φ, 2R sin θ sin φ, 3R cos θ sin φ > < R 2 cos θ sin 2 φ, R 2 sin θ sin 2 φ, R 2 cos φ sin φ > [R 4 cos 2 θ sin 3 φ cos φ 2R 3 sin 2 θ sin 3 φ + 3R 3 cos θ cos φ sin 2 φ]dφdθ the first and last terms go to zero because of cos φ being odd about the φ π/2 axis, and the middle term gives, with R 2 here for this sphere, 2π π 2 4 sin 2 θ sin 3 φ dφdθ... 64π/3. (The way to integrate this would be to use a half angle formula for sin 2 θ and then to rewrite sin 3 φ sin φ(1 cos 2 φ) and use a substitution w cos φ.) 2) hortcut Way: ince we know that d sphere R 2 sin φdθdφ, we can just find F ˆn first. Here, we get n first by taking the gradient of the level surface that is the sphere, x 2 + y 2 + z 2 R 2, in other words, n (x 2 + y 2 + z 2 ) < 2x, 2y, 2z >. Then after we normalize it, we have ˆn < 2x, 2y, 2z > < x, y, z > 4(x 2 + y 2 + z 2 ) x 2 + y 2 + z 1 < x, y, z > on the sphere of radius R. 2 R 11
o, taking dot product, F ˆn < xz, 2y, 3x > 1 R < x, y, z > 1 R (x2 z 2y 2 + 3xz). Now we know that x R cos θ sin φ, y R sin θ sin φ, z R cos φ and also that d R 2 sin φdθdφ so F d F ˆnd 2π π 2π π 1 R (R3 cos 2 θ sin 2 φ cos φ 2R 2 sin 2 θ sin 2 φ + 3R 2 cos θ sin θ sin φ) R 2 sin φdφdθ (R 4 cos 2 θ sin 3 φ cos φ 2R 3 sin 2 θ sin 3 φ + 3R 3 cos θ sin θ sin 2 φ)dφdθ which is the same integral that resulted from the standard way, again, gets 64π/3. (After some clever integration, and plugging in R 2). #3 A more standard example. First, we parameterize the paraboloid. ince z is between the paraboloid and the plane z 1, we have that r(x, y) < x, y, x 2 + z 2 >, on x 2 + y 2 1. Then, for examples, when we take the partials and cross them, we have r y <, 1, 2y >, r x < 1,, 2x >, r y r x < 2x, 2y, 1 > (CAREF UL). However, we want to have upward orientation, so the z-component of this normal vector must be positive, not negative (here, it is 1). Therefore, we should actually take r x r y < 2x, 2y, 1 > for the correct orientation. Using this now, we have that by definition, F d F(r(x, y)) r x r y dxdy x 2 +y 2 1 < x 2, xy, x 2 + y 2 > < 2x, 2y, 1 > dxdy x 2 +y 2 1 ( 2x 3 2xy 2 + x 2 + y 2 )dxdy P olar 2π 1 in which the final answer is π/2. x 2 +y 2 1 x 2 +y 2 1 [ 2x(x 2 + y 2 ) + x 2 + y 2 ]dxdy ( 2r cos θ r 2 + r 2 )rdrdθ 2π 1 ( 2r 4 cos θ + r 3 )drdθ 12
#37 on t fear, this is just fundamental theorem of line integrals! In general, if a curve looks really bad, it may be an implied hint to use the fundamental theorem. First, we d need to find f such that F f. You should get that F f for the function f(x, y, z) x 3 yz 3xy + z 2 + K. Then, by the fundamental theorem, F dr f(final) f(initial) f(, 3, ) f(,, 2) + K 4 K 4. C If you need help to get f, here s the solution: 1) Integrate any component (I ll just pick the z component) and add in the appropriate constant function of integration to get an initial form of f. f(x, y, z) g(x, y) + (x 3 y + 2z)dz x 3 yz + z 2 + g(x, y). 2) olve for g(x, y) by making f x and f y match the x and y components of the vector field F. f x 3x 2 yz + g x needs to equal 3x 2 yz 3y. Hence, g x 3y so then g(x, y) h(y) + 3ydx h(y) 3xy. o, from analyzing f x, we see f(x, y, z) x 3 yz + z 2 3xy + h(y). f y x 3 z 3x + h y (y) This means h y (y) so thus h(y) K constant. We see needs to equal x 3 z 3x. f(x, y, z) x 3 yz + z 2 3xy + K. 13