Math 3B iscussion ession Week 1 Notes March 14 and March 16, 17 We ll use this week to review for the final exam. For the most part this will be driven by your questions, and I ve included a practice final exam below as a starting point. After each problem you ll see (olution); clicking this will take you to the relevant solution. This exam is a modified version of a final from tanford s Math 5 course, and several others like it can be found at http://math.stanford.edu/~ww/5.php?contente. Because tanford s Math 5 syllabus isn t necessarily exactly the same as ULA s Math 3B syllabus, some of the emphasis and terminology in these exams might be different from ours; use your best discretion. Finally, you should be aware that this is just a sampling of problems that I think you should know how to do, and this should not be considered an indication of the problems you ll see on the exam. In case they re of any help, I ve also reproduced the two tables we made in recent weeks, minus some of the explanations/interpretations. 1
Throughout these tables, T represents a positively-oriented unit tangent vector to a curve, n represents a positively-oriented unit normal vector to a surface, and N(u, v) represents a positivelyoriented normal vector whose length depends on our parametrization of the surface. arc length/surface area integral of scalar function f integral of vector field F fluid flowing through or Line Integrals ˆ b a ˆ b a r (t) dt f(r(t)) r (t) dt ˆ ˆ F dr : (F T)ds ˆ F nds ˆ b a ˆ b a F(r(t)) r (t)dt F(r(t)) n(t) r (t) dt urface Integrals G u G v dudv f(g(u, v)) G u G v dudv F d : (F n)d F(G(u, v)) N(u, v)dudv F d F(G(u, v)) N(u, v)dudv FTs Integral across boundary Integral of derivative across solid 31A version, f : R R line integrals, a path from P to Q Green s Theorem tokes Theorem Green s, flux version ivergence Theorem f(b) f(a) f(q) f(p ) F dr F dr (F n)ds W F d ˆ b a ˆ f (x)dx f dr ( F x F ) 1 da curl z (F)dA y curl(f) d ( F1 x + F ) da div(f)da y W div(f) d
1. (a) ketch the region R of integration in the following double integral. x xe y5 dydx (b) Express the region R as a horizontally simple region. (c) Evaluate the integral by changing the order of integration. (olution). Given a solid region W R 3, the moment of inertia of W around an axis (line) l R 3 is given by M l d l dv, where d l : R 3 [, ) is the function which gives the distance of a point in R 3 from the axis l. W (a) Let T be a solid cone whose base is the disk x + y 1 in the xy-plane and whose vertex is the point (,, ). et up (but do not evaluate) a triple integral in cylindrical coordinates which computes the moment of inertia of T around the x-axis. (b) Let T be the solid ball of radius 1 centered at (,, ). et up (but do not evaluate) a triple integral in spherical coordinates which computes the moment of inertia of T around the z-axis. (Hint: First consider the change of variables u x, v y, w z.) (olution) 3. onsider the curve parametrized by r(t) (e t, 1 3 e3t + e t ) for t 1. (a) Find the arclength of. (b) Find the x-coordinate of the centroid of. (c) Let be the surface obtained by rotating around the line x. Find the area of. (olution) 4. (a) onsider the change of variables u x 4 + y, v y x. Find the Jacobian (x, y) (u, v). (b) Find the area of the region in the first quadrant enclosed by the ellipses x 4 + y 1 and x 4 + y 4 and the lines y x and y 4x. (olution) 5. Let T be the solid enclosed by the planes z x+y and x+y and the paraboloid z y. et up (but do not evaluate) a triple integral in rectangular coordinates which computes the volume of T by 3
(a) regarding T as x-simple (i.e., make x the innermost integral); (b) regarding T as z-simple. (olution) 6. Let be the curve parametrized by r(t) (t, sin t) for t π/. (a) ompute ydx. (b) Let R be the region in the plane bounded by the y-axis, the line y 1, and the curve. Use Green s theorem to find the area of R. (olution) 7. Let be the portion of the cylinder x + y x which lies above the xy-plane and below the surface z x. (a) Write down a parametrization of. Be sure to specify the domain. (b) Find the area of. (olution) 8. Let be the surface traced out by G(u, v) ( [, 1] in the uv-plane. v, u, uv) over the domain [, 1] (a) Find the area of. (Warning: The map G is not one-to-one on. 1 ) (b) Find the x-coordinate of the centroid of. (olution) 9. Let be the portion of the sphere x + y + z 4 which lies above the plane z, and let F(x, y, z) (x 3y + z, 3x + y + z, x + y 3z). ompute the flux of F through in the outward-pointing direction. (olution) 1. Let T be the triangle with vertices A (1, 1, 1), B (3,, ), and (1,, ), parametrized in that order. Let F(x, y, z) (4z, x, y ). (a) Find the area of the triangle. (b) Find the equation of the plane in which the triangle lies. (c) ompute curl(f). (d) ompute F dr. (olution) T 11. Let F(x, y, z) (xy + z, yz + x, zx + y ). 1 I m not entirely convinced that this extra layer of difficulty was intentional on the part of the test-writer; I think the domain may be a typo. If you can t figure out how to deal with the lack of one-to-one-ness, maybe skip down to the solution, find the domain we really want, and then see if you can do the rest of the problem on your own. 4
(a) Find a potential for F. (b) Let be the curve parametrized by r(t) (cos t, sin t, t) for t π. F dr. (olution) ompute 1. Let T be the solid region which lies below the surfaces z 1 x and z 1 y and above the xy-plane. (a) Find the volume of T. (b) Find the outward flux through the boundary of T of the vector field F(x, y, z) (zx, z y, 3z z x). (olution) 5
olutions 1. (a) The region of interest here is the region where x 1 and x y 1, which we can easily plot: (b) Expressing R as a horizontally simple region means making slices which are parallel to the x-axis, such as the red slice in the above plot. The possible y-values at which we can make such a slice range from to 1, and once y is fixed, x is allowed to vary between and y. o R {(x, y) y 1, x y } gives a horizontally simple description of R. (c) According to Fubini s theorem we have x xe y5 dydx ˆ y xe y5 dxdy [ x ey5 ] y dy 1 y 4 e y5 dy. We may then let u y 5, so that du 5y 4 dy. Notice that our bounds of integration do not change. Then we have x xe y5 dydx 1 1. (a) Here s a plot of the cone we re interested in: e u du e 1 1. In cylindrical coordinates our ranges for r and θ are obvious: we let r vary between and 1, while θ varies between and π. The bounds for z require only a little more work. We know that we should have z, and also that the upper bound for z should depend only on r, and not on θ. In fact, we can write the relationship between z and r as Az + Br 1 for some choices of A and B, since the upper bound for z depends on r in a linear fashion. We know that when r our upper bound for z is, so A 1/. 6
When z, our upper bound for r is 1, so B 1. o we have z r. Altogether, the cone is described by θ π, r 1, z r. For the x-axis we re interested in integrating d x y + z, the square of the distance of a point from the x-axis, so we have ˆ π ˆ r I x (y + z )dv (r sin θ + z )rdzdrdθ. T In case you prefer to first describe the cone in rectangular coordinates and then make substitutions, we can do that too. In rectangular coordinates we can desribe T with the inequalities x + y 1 and z x + y. The second inequality may not be so obvious, but consider this. The standard (upwardopening) cone is described by z x + y. We want to turn this upside down and make it sharper, so we have z λ x + y for some λ >. We also need to shift the cone up so that the vertex is at (,, ), so we in fact have z λ x + y. ince we want z whenever x + y 1, we have λ. These two inequalities of course translate to r 1 and z r, and then θ is allowed to fill its usual range from to π. (b) In rectangular coordinates T is described by (x ) + y + z 1. We want to use the transformation indicated by the hint, but we remember that our transformation should map into the region T over which we want to integrate, so we write x u +, y v, z w, and then define G: R 3 R 3 by Now we have I z T d zdv G(u, v, w) (u +, v, w). G (T ) We can quickly check that Jac(G) 1. Also, (d z G(u, v, w)) Jac(G) dudvdw. G (T ) {(u, v, w) u + v + w 1} so we need to integrate d z G (u + ) + v over the unit ball. We now introduce spherical coordinates: u ρ sin φ cos θ, v ρ sin φ sin θ, w ρ cos φ, which then gives dudvdw ρ sin φdρdφdθ. o the moment of inertia is given by I z ˆ π ˆ π ((ρ sin φ cos θ + ) + ρ sin φ sin θ)ρ sin φdρdφdθ. 7
3. (a) From the given parametrization we compute r (t) e t, e 3t e t, so r (t) 4e t + e 6t e t + e t e 6t + e t + e t (e 3t + e t ) e 3t + e t. Then the arclength of is given by ˆ 1ds r (t) dt e 3t + e t dt [ 1 3 e3t e t ] 1 ( ) e 3 3 e ( ) 1 3 1 1 3 (e3 3e + ). (b) We never really talked about centroids of curves, but this has the formula you would expect: x M ˆ y M, where M y xds. Here M is the mass of the curve, which is the same as the arclength if we assume constant density 1. Now let s compute M y : ˆ [ ] 1 1 M y xds (e t ) r (t) dt e 4t + 1dt 4 e4t + t ( ) ( ) e 4 1 4 + 1 4 + 1 (e4 + 3). Then x M 1 y M (e4 + 3) 1 3 (e3 3e + ). (c) For this part it will probably help to see the curve. On the left we have a plot of (in blue), along with the line x (in orange). On the right we see the surface which results from rotating around the line x. If we fix a value t 1 to get a point on, the rotation around the line x will turn this point into a circle on, such as the green one seen in the figure. We can compute the circumference of this circle if we know its radius, and this value will be the length of the purple radial line in the first plot. Once we know the circumference of the circle generated by a point on, we can integrate across to get the surface area of. In pseudo-math, ˆ ˆ surface area circumference ds π radius ds π R(t) r (t) dt. 8
The last integral is the thing we want to compute; we just need to write down R(t), the radius at time t. This is the length of the purple line above, and should be equal to the x-value of r(t), plus the one unit to get from the y-axis to the line x. That is, o R(t) e t + 1. Area π (e t + 1)(e 3t + e t )dt π e t (e 3t + e t )dt + π [ e 4 + 3 π + e3 3e ] + π 3 3 (3e4 + e 3 + 13 6e ). The first integral above was computed in part (b), the second in part (a). 4. (a) We have so (x, y) (u, v) (u, v) (x, y) ( ) (u, v) (x, y) ( ) x/ y det y/x 1 1/x + y x, 1 1 + y x (b) The region R in which we re interested is described by 1 + 4y /x 1 + 4v. R {(x, y) 1 x 4 + y 4, y/x 4}. This is the preimage under our transformation of the region so Area R ˆ 4 {(u, v) 1 u 4, v 4}, 1dA Here s a plot of the region R: (x, y) (u, v) da ˆ 4 ˆ 4 1 1 + 4v dudv 6 1 + 4v [3 arctan(v)]4 3 arctan(8) 3 arctan(4). e 3t + e t dt 9
5. (a) ince the innermost integral is in terms of x, we want to bound x between expressions that depend on y and z. We can rewrite the bound z x + y as z y x, and rewrite the bound x + y as x y. o we have z y x y. For this set of bounds to make sense, we must have z y y, so z y. ombining this with the requirement that y z, we have y z y. Finally, we need y y, so y + y. That is, (y + )(y 1), so y 1. The volume of T is then given by vol(t ) ˆ y ˆ y z y y 1dydzdx. (b) We now want to rewrite our bounds so that z lives between expressions depending on x and y. The bounds z x + y and y z immediately give y z x + y. This set of bounds then requires y x+y, so y y x. We also know that x+y, so x y and we see that y y x y. Finally, y y y tells us that y + y, so again we have y 1. o we can also compute the volume of T as vol(t ) 6. (a) Notice that r (t) t, cos t, so ˆ ˆ ydx y, dr ˆ x+y ˆ y y ˆ π/ y y 1dydxdz. sin t, r (t)dt ˆ π/ t sin tdt. We can use integration by parts, letting u t and letting dv sin tdt. Then ˆ ˆ π ydx [ t cos t] π/ ( cos u)du [ sin u] π/. (b) Here s a plot of the region R: 1
We re asked to compute the area of R using Green s theorem, which we recall can be done using a number of different integrals: Area xdy ydx 1 xdy ydx. R R R ince we ve already computed ydx in part (a), we ll use the middle option. If we call the top part of R (where y 1) 1 and call the left part (the y-axis), we have (ˆ ˆ ˆ ) Area ydx ydx + ydx + ydx R 1 ( ) + ˆ ( π 4 π /4 1dx + ) π 4. We have ydx because x is constant along. 7. (a) Our cylinder might be more readily parametrized if we rewrite its defining equation to look more familiar. The equation x + y x can be rewritten x x + y x x + 1 + y 1 (x 1) + y 1. o this is the standard unit-radius cylinder, translated 1 unit in the x-direction. Inspired by cylindrical coordinates, we give the parametrization G(u, v) (cos u + 1, sin u, v). Notice that u and v play the roles usually played by θ and z, respectively, and that the x-coordinate has been bumped up by 1 unit. We also must be clear about the domain of G. ince is bounded below by the xy-plane and above by z x, we have For u and v this means z x. v (cos u + 1). Notice that the requirement that (cos u + 1) imposes no condition on u, so u π. (b) Using the above parametrization, we have so o G u G v 1, and thus surface area G u sin u, cos u, and G v,, 1, i j k G u G v sin u cos u cos u, sin u,. 1 ˆ π ˆ (cos u+1) G u G v dudv 11 ˆ π ˆ (cos u+1) 1dudv 3π.
8. (a) We noticed that G is not one-to-one on since, for example, ( 1 G, 1 ) ( G 1 ),. If we restrict our attention to the subdomain [, 1] [, 1], G will be one-to-one on the interior of, but will also trace out the same surface as before. We can verify this by explicitly constructing an inverse map: G (x, y, z) ( z x, x ) This inverse map only makes sense on the portion of where x >, but this corresponds precisely with the interior of, where v >. Whether or not we were really supposed to care about these domain scruples is unclear to me, but in any case the domain we care about is [, 1] [, 1]. Once we ve restricted to this domain, we have Area() 1d G u G v dudv. We compute so G u, u, v and G v v,, u, i j k G u G v u v v u u, v, uv.. Then G u G v u 4 + v 4 + 4u v ( u + v ) (u + v ). o the surface area is Area() (u + v )dudv [ u 3 (v + /3)dv [ 3 v3 + 3 v (b) For the x-coordinate of the centroid, we need to compute M yz xd, 3 + uv ] 1 ] 1 dv 4 3. and then we ll have x M yz /M. Here M is the mass of the surface, which we can take to be the area if we assume uniform density 1. Under the parametrization, x so we have ˆ 1 ( ) M yz v G u G v dudv v (u + v ) dudv 1 v,
Finally, v (u + v )dudv v (/3 + v )dv [ 9 v3 + 5 v5 [ u v 3 3 + uv ] 1 x 8/45 4 /3 7 15. ] 1 dv 9 + 5 8 45. Here s a plot of the surface, with the centroid as a red dot. The strange point of view here was chosen so that we could see that the centroid is a point somewhere in the convex hull of our surface, but not actually on. 9. Recall that this flux is computed as a vector surface integral: F d. At this point we could parametrize using our knowledge of spherical coordinates, then compute a normal vector for this parametrization and end up with a big, ugly integral. Instead, let s use the divergence theorem. Our surface is a portion of a sphere, and it intersects the plane z in a circle. We denote by P the solid disk in the plane z which is bounded by this intersection. Then and P together bound a three-dimensional region W, as seen here: 13
We recall from the divergence theorem that div(f)dv F d W W F d + F d. P ince div(f) 1 + 3, the integral on the left vanishes and we have F d F d. o instead of integrating F over, we only have to integrate F over P. Now the intersection of and P is the circle where x + y + z 4 and z, so x + y 3. That is, P is a disk of radius 3 centered at (,, ) and contained in the plane z. o we can parametrize P by G(u, v) (u cos v, u sin v, ), where u 3 and v π. The tangent vectors for this parametrization are given by G u cos v, sin v, and G v u sin v, u cos v,, so i j k G u G v cos v sin v,, u. u sin v u cos v Now we have to be a little bit careful. We want our normal vector N(u, v) to be pointing out of the region W, which means that the z-coordinate should be negative along P. ince u 3, we take N(u, v) G u G v. Finally we notice that so F(G(u, v)) u cos v 3u sin v, 3u cos v + u sin v 1, u cos v + u sin v + 3, F(G(u, v) N(u, v) (u cos v + u sin v + 3)( u) (u cos v + u sin v + 3u). We re now ready to integrate: P ˆ π F d ˆ π ˆ 3 P ˆ π [ u cos v + u sin v + 3udu 3 u3 cos v + 1 3 u3 sin v + 3 ] 3 u dv 3 cos v + 3 sin v + 9 dv [ 3 sin v 3 cos v + 9 v ] π ( 3 + 9π ) ( + ) 3 + 9π. o the total outward flux of F through is F d P F d 9π. We could also have done this problem using tokes theorem instead of the divergence theorem. The trick here is to find a vector field G so that curl(g) F. This would give us F d curl(g) d G dr. 14
One such G is given by G(x, y, z) 3yz 3xz xy, 3yz + xy, xy + yz xz. (You can verify that curl(g) F.) Then we can parametrize by r(t) ( 3 cos t, 3 sin t, ), t pi and compute F d ˆ π G(r(t)) r (t)dt 9π. I would not recommend this second approach, though, since finding the vector field G can be a messy process. 1. (a) The triangle is spanned by the vectors u B A,, and v A,, 1. We can compute the area of T as 1 u v. We have i j k u v 4,, 4, 1 so Area(T ) 1 16 + 4 + 16 3. (b) The vector u v above gives us a normal vector to the plane P containing T, so we can write 4x y 4z λ for every point (x, y, z) in this plane, for some constant λ. Equivalently, we can write 4x + y + 4z λ. ince we know that, say, (1, 1, 1) is a point in this plane, we have 4(1) + (1) + 4(1) 1 λ, so P is described by 4x + y + 4z 1, or x + y + z 5. (c) We have i j k curl(f) F x y z 4z x y y, 8z, 4x y, 4z, x. 15
(d) Let denote the solid triangle contained in P whose boundary is T. theorem tells us that F dr curl(f) d. T Then tokes At first it seems that we ll need to parametrize in order to compute the integral on the right, but remember that vector surface integrals are defined by G d (G n)d, where n is a unit normal vector to. We can easily obtain n: n 4,, 4 4,, 4 1 4,, 4. 6 o we have T F dr curl(f) d 1 y, 8z, 4x 4,, 4 d 6 8 + z + x))d 6 (y 8 5d 6 8 6 5 Area() 4 6 3. 11. (a) We re looking for a function f so that f x xy + z, f y yz + x, and f z zx + y. By integrating the first of these conditions we see that f has the form f(x, y, z) x y + xz + g(y, z) (1) for some function g of y and z. We can differentiate this equation with respect to y to see that f y x + g y, so g y yz. Integrating this with respect to y, we have g(y, z) y z + h(z) for some function h of z. Finally, we differentiate (1) with respect to z to find that xz + g z xz + y + h (z). ince we know that f z zx + y, we see that h (z), so h is a constant function. o g(y, z) y z and we see that f(x, y, z) x y + xz + y z is a function with the property that f F. 16
(b) ince F has a potential function, this vector field is conservative, so we may apply the fundamental theorem of conservative vector fields. In particular, ˆ F dr f(r(π)) f(r()) f(,, π) f(1,, ) π π. 1. The points (x, y, z) in T must simultaneously satisfy the inequalities z 1 x and z 1 y, so we can write our region as z min(1 x, 1 y ). When x y we have 1 x 1 y, and when x y we have 1 y 1 x. o T is the region between the xy-plane and the plot of the function { 1 x f(x, y), x y 1 y, x y over the region [, 1] [, 1]. We choose this region because this is precisely the region where f(x, y). (a) With the work done above, we see that the volume of T is given by vol(t ) f(x, y)da (1 x )da + 1 (1 y )da, where 1 {(x, y) x y } and {(x, y) x y }. We an describe 1 and in terms of inequalities: { 1 (x, y) x 1 } { and x y x (x, y) y 1 }. y x y o we have vol(t ) ˆ x x (1 x )dydx + x (1 x )dx + ˆ y y (1 y )dxdy y (1 y )dy 4 u (1 u )du. The last equality simply relies on the fact that variable names don t matter. We integrate absolute value piecewise: u (1 u )du ˆ u(1 u )du + [ 1 u + 1 4 u4 ] + ( 1 1 ) ( 1 + 4 1 4 Altogether, vol(t ) 4 u (1 u )du. Here s a plot of the region T : 17 u(1 u )du [ 1 u 1 4 u4 ) 1. ] 1
(b) We can use the divergence theorem to write F d T T div(f)dv. ince div(f) zx 1 + 3 xz, the integral on the right will be significantly easier to compute. Indeed, F d dv vol(t ) 4. T o the total flux of F through T is 4. T 18