Calculus with Analytic Geometry 3 Fall 2018

Transcription

1 alculus with Analytic Geometry 3 Fall 8 Practice Exercises for the Final Exam Solutions. The points P (,, 3), Q(,, 6), and (4,, 4) are 3 vertices of the parallelogram spanned by the vectors P Q, P. (a) Find the coordinates of the fourth vertex. (b) A bug is crawling along the line of equation r(t) = ( + t)i + ( + t)j + (3 + t)k. Suppose time is measured in seconds, distance in meters. At time t = it is at the point of coordinates (,, 3). It then moves in the positive t direction for a distance of exactly meter along the line, and stops. Determine the coordinates of the point Z at which the bug has stopped. (c) Find the volume of the parallelepiped having the parallelogram of before as a base, and Z as a vertex. The fourth vertex of the parallelogram, call it S, will satisfy P S = P Q + P =,, 3 + 3,, = 3,, 4. Since P S is obtained by subtracting the coordinates of P from those of S; to get the coordinates of S we add those of P to the components of P S. The answer to part (a) is (4,, 7). For part (b) we have to figure out first for which t > is r(t) (,, 3) =. This works out to t + t + 4t =, thus t = / ( 6. The answer is thus Z = +, +, 3 + ) For (c) we have that the parallelepiped is spanned by the vectors P Q, P, and vecp Z = (/ 6,, 6, / 6). The volume is then the absolute value of the triple product 3 ( P Q P ) P Z = 3 = 9 / 6 / 6 / 6 6 The answer is onsider the curve given parametrically by x = a cos t, y = b sin t, z = c sin t, where a, b, c are positive real numbers. Show that it lies in a plane. Find the equation of the plane. One can solve this by some educated guessing or in a more systematic way. I begin with the systematic way. A plane is determined by three points. So if we find three non-collinear points on the curve they will determine a plane and if the curve lies in a plane, that should be the plane. So for the three points let us take some values of t where it is easy to compute the curve coordinates for that t. I think the best values are t =, π/4, and π/. With t = we get the point (a,, ) (take as base point), with t = π/4, the point (a/, b/, c/ ). With t = π/ we get (, b, c). A vector perpendicular to the plane through these three points is thus given by i j k a a b c a b c = ( )acj + ( )abk. Any vector parallel to this one would do; one such vector is cj+bk. The equation of the plane perpendicular to this vector and through (a,, ) is cy + bz =. Plugging the expression for the curve into the equation of the plane we get cb sin t + bc sin t =

2 which clearly holds for all t. The answer is cy + bz =. There is an alternative approach. The vector, c, b is easy to guess as being perpendicular to all values of the curve, meaning the curve lies on the plane through the origin perpendicular to this vector. 3. A curve is described by the vector function r(t) = t i + t 3 j + t k (t > ). Find the parametric equations of the tangent line at the point where t =. We have r (t) = ti + 3t j t k. For t =, the curve is at r() = 4i + 8j + k and a tangent vector is r () = 4i + j 4k. Thus, the following are parametric equations for this line. x = 4 + 4t, y = 8 + t, z = 4 t. 4. A curve is described by the vector function r(t) = t i + t 3 j + t k (t > ). Find the equations of the normal plane at the point where t =. In the previous exercise we found that the point where t = is (4, 8, /) and that direction vector of the tangent line is 4,, /4. This allows one to write out at once the equation of the plane as 4(x 4) + (y 8) 4 (z ) =. One can work on this equation and rewrite it in the form 6x + 48y z = A curve is given parametrically by x = t, y = t t 3, z =. Show that it self-intersects (goes twice through the same point) and find the equations of the two tangent lines at that point. For the curve to self intersect there must be two parameter values, say s < t at which all three coordinates are equal. This gives the following equations for s, t: s = t, s s 3 = t t 3, =. The last equation is, of course, always true and can be ignored. From the first one we get s = ±t thus t = s. Plugging this into the second equation gives s s 3 = s + s 3, thus (s + s 3 ) =. We get three solutions: s =,,. If s = then t = s =. If s = then t = ; but t > s. Similarly, if s =, then t = violating s < t. The solutions are s =, t =. The point in question is the end point of the vector r( ) = r() =,, ; i.e., the point (,, ). Now r (t) = t, 3t, so two tangent vectors directing the two lines are r ( ) =,,, r () =,,. From all of this information we get at once that the two lines have equations given parametrically by x = t, y = t, z =, and x = + t, y = t, z =. 6. (a) Find the length of the curve described by the position vector r(t) = ti + 8t 3/ j + 3t k for t. It might help to realize that 44 = 4 36 and that (t + ) = t + 4t + 4. r (t) = i + t / j + 6tk, thus r (t) = t + 36t = 36(4 + 4t + t = 36(t + ) = 6(t + ). Thus L = 6 (t + ) dt = 5.

3 3 (b) Suppose that the curve of part (a); that is, the curve of position vector r(t) = ti + 8t 3/ j + 3t k, gives the position of a particle at time t. Determine the time t h at which the particle has gone through half the length of the curve. That is, find t h such that the length of r(t), t t h equals the length of r(t), t h t. Since the total length of the curve is 5 (length units) we must find t h such that the length of the segment for t t h is 5/ units. This means that we should have th r (t) dt = 5. Using the expression for r (t) found in part (a), we get 6 th (t + ) dt = 5 ; that is, 3t h + t h = 5. Multiplying by, we get 6t h + 4t h 5 =. Solving we get t h = 4 ± 936 Since < t h < we have to choose the plus sign, getting = ± 78. t h = Find the vectors T, N, and B for the curve given by r(t) = t, 3 t3, t at the point where t =. Note: omputations can be a bit nasty, so work carefully. There is a point where it could help to know that (t + ) = 4t + 4t +. We have r (t) = t, t,, thus r (t) = 4t + 4t 4 + = (t + ) = t +. Thus Setting t = we get that t T(t) = t +, t t +, t +. T = 4 9, 8 9, 9. Next, being careful!, T (t) = (t + ) 8t (t + ), 4t(t + ) 8t 3 4t (t + ), (t + ) = 4t (t + ), 4t (t + ), 4t (t + ). Since we do not have to differentiate any more, we could set t = right now. Makes life easier. But to make things as challenging as possible, we won t do this yet. Then Thus T (t) = = ( 4t (t + ) ) + 6t + 6t 6t4 (t + ) + 6t + 4 = (t + ). N(t) = t t +, t t +, t t +.

4 4 Setting t = we get N = 7 9, 4 9, 4 9. Finally B = T N = 4 9, 9, Find the velocity, acceleration, and speed (as functions of time t) of a particle with the position vector r(t) = t sin t, t cos t, t. v(t) = r (t) sin t + t cos t, cos t t sin t, t, speed = v(t) = (sin t + t cos t) + (cos t t sin t) + 4t = + 5t, a(t) = v (t) = cos t t sin t, sin t t cos t,. 9. Find the position and velocity vectors of a particle if its acceleration and initial velocity and position are a(t) = i + j + tk, v() =, r() = i + k. We have v(t) = a(t) dt = (i + j + tk) dt = ti + tj + t k +. Setting t = and equating with v() =, we get =. Thus v(t) = ti + tj + t k. Next r(t) = v(t) dt = (ti + tj + t k) dt = t i + t j + 3 t3 k + D. Setting t = and equating to r() = i + k, we get D = i + k, thus r(t) =. Let g(s, t) = f(u(s, t), v(s, t)) where f, u, v are differentiable, u(, ) =, v(, ) = 3, u s (, ) = 5, v s (, ) = 7, f(, ) =, f(, 3) = 4, f(5, 7) =, f u (, 3) =, f v (, 3) =, f u (, ) =, f v (, ) =. ( ) ( ) t + i + t j + 3 t3 + k. Determine g s (, ). Some of the information provided is useless and only included to see if you can select what you need. setting s =, t = we get g s (s, t) = f u (u(s, t), v(s, t))u s (s, t) + f v (u(s, t), v(s, t))v s (s, t); g s (, ) = f u (, 3)u s (, ) + f v (, 3)v s (, ) = ( ) = 4.. Find an equation of the tangent plane to the given surfaces at the specified point. (a) z = x + y 3 at the point where (x, y) = (, ); that is at (,, 9). z x = x, z y = 3y so that z x (, ) =, z y (, ) = 4. The equation is z = 9 + (x ) + (y ).

5 5 (b) xy + yz + zx = 3 at (,, ). The surface is given in the form f(x, y, z) = 3 where f(x, y, z) = xy + yz + zx. Now f(,, ) = 3, 3, 3, thus the equation is 3(x ) + 3(y ) + 3(z ) =. This equation reduces to x + y + z = 3. (c) x + ze y + ye z = 4 at the point where (x, y) = (, ). No value of z is given, so it has to be figured out. If x =, y = the equation becomes + z = 4, thus z = 3. The point at which the plane is tangent is (,, 3). If f(x, y, z) = x + ze y + ye z, then f(,, 3) =, 3 + e 3,. The equation is x + (3 + e 3 )y + z 3 =, or x + (3 + e 3 )y + z = 4.. Let f(x, y) = x + 4xy y. (a) Find the unit vectors giving the direction of steepest ascent and steepest descent at (, ). (b) Find a vector that points in a direction of no change at (, ). (a) f(x, y) = x + 4y, 4x y ; f(, ) = 8, 6. Now f(, ) = =. The unit vector in the direction of steepest ascent is 4 5, 3 5., of steepest descent, 4 5, 3 5. For (b), both 3 5, 4 5 or 3 5, 4 point in directions of no change; either one can be used to answer part (b) A mountain having an elliptical base can be described by the equation z = 5 x 4y. A climber will try to reach the top starting at the point of coordinates (3,, ). The climber wants to be always going in the direction of steepest ascent. The climber decides to follow the path that can be parameterized by x = 3e t, y = e 4t, z = 5 9e t 6e 8t, t. (a) Show that this is a path that is indeed on the mountain. so it is a mountain path. 5 x(t) 4y(t) = 5 9e t 6e 8t = z(t), (b) Show that if the climber follows it, at each point the climber will be moving in the direction of steepest ascent. Let P (x, y, z) be a point on the path, so for some value of t we have x = 3e t, y = e 4t, z = 5 9e t 6e 8t. Since the climber is moving along the path in question, the climber s direction is given by the tangent to the curve, namely T = 3e t, 8e 4t, 8e t + 8e 4t. The last component is not too important here; it is positive showing the motion is upward, but that s all. The two first components; that is, u = 3e t, 8e 4t, give the direction of motion. The direction of steepest ascent is given by the gradient; z = x, 8y. At the point in question we have x, 8y = (3e t ), 8(e 4t ) = 6e t, 6e 4t = u. This shows that the vector governing the direction of motion is parallel to the direction of steepest ascent (and pointing in the same direction), so the path is indeed a path of steepest ascent. 4. Find the critical points and use the second derivative test to classify them of f(x, y) = x 4 +y 4 4x 3y +. To find the critical points, we set the first order partials to. f y = 4y 3 3 =. f x = 4x 3 4 =, There is a single solution, (x, y) = (, ). This is the only critical point. Now D(x, y) = f xx(x, y) f xy (x, y) f xy (x, y) f yy (x, y) = x 3 y 3

6 6 so that D(, ) = 96 = 5 >. Because D(, ) is positive, we have a relative minimum or maximum at (, ). Because f xx (, ) <, it is a relative minimum. 5. Find the critical points and use the second derivative test to classify them of f(x, y) = xye x y. We set the first order partial derivatives to : f x = ye x y xye x y = ( x)ye x y =, f y = xe x y xye x y = ( y)xe x y =. The first equation is satisfied if (and only if) x = or y =. Going to the second equation with x =, the only solution is y =. So (, ) is a critical point, and the only one with first component. If we put y = in the second equation, we see that we must have x =. So (, ) is a critical point; the only one with second component. The critical points are (, ) and (, ). The Hessian determinant is D(x, y) = y(x )e x y ( x)( y)e x y ( x)( y)e x y x(y )e x y. From this D(, ) =, so (, ) is a saddle point. Next, D(, ) = e 4 >. so we have a relative extremum. Since f xx (, ) = e <, (, ) is a relative maximum. 6. Find ALL critical points of f(x, y) = 3x x y 4y + y 3 and classify each one of them as local maximum, local minimum, or saddle point. We compute the partial derivatives and set them to zero: f x : 6x 4xy =, f y : x 8y + 6y =. The first equation has one solution x =. Using this in the second equation we get 8y + 6y =, thus y(6y 8) =, which implies y = or y = 4/3. Two critical points are (, ) and (, 4/3). If x then the first equation has the only solution y = 3/. Using this in the second equation gives x = ± 3/. The critical points are (, ), (, 4/3), ( 3/, 3/), ( 3/, 3/). For the classification we compute the Hessian determinant D(x, y) = 6 4y 4x 4x y 8 = 4y 48y 48 6x. Now for the classification. D(, ) = 48 <, (, ) is a saddle point. D(, 4/3) = 6/3 >,, f xx (, 4/3) = /3 >, (, 4/3) is a relative minimum. D(± 3/, 3/) = <, ( 5/, 3/) and ( 5/, 3/) are saddle points. 7. Show that the second derivative test is inconclusive when applied to f(x, y) = x y 3 at (, ). Describe the behavior of the function at the critical point. f x (x, y) = xy, f y (x, y) = x, so f x (, ) = f y (, ) =. The point (, ) is indeed critical. We also have f xx (, ) = y (,) =, f xy (, ) = x (,) =, f yy (, ) =, so of course the Hessian determinant is and the test is inconclusive. Now f(, ) = 3. To describe the behavior at (, ) we might notice that if x >, then xy > so that as we move away from (, ) along the positive x axis, the function grows. But for the same reason, moving away from (, ) along the negative x-axis, the function decreases. This means we can t have a relative maximum or minimum at (, ). It follows that there is a saddle point at (, ).

7 7 8. Show that the second derivative test is inconclusive when applied to f(x, y) = sin(x y ) at (, ). Describe the behavior of the function at the critical point. I will begin describing the behavior of the function at the point. It is clear the function has a relative minimum at (, ). In fact, if both x, y are different from and small, then sin(x y ) >. If one of x or y is, then sin(x y ) =. In either case sin(x y ) = sin( ), so there is a relative minimum at (, ). One sees that f x, f y, f xx, f xy, f yy are all at(, ), so it is a critical point (as it had to be, being a relative minimum) and the Hessian determinant is. For the next few exercises things to know are:. In a closed and bounded region, a continuous function will assume a maximum value and it will assume a minimum value.. These values have to be assumed either at a critical interior point or on the boundary. They cannot be assumed anywhere else. Maybe I should add a third thing, if a problem asks for the maximum or minimum VALUE of a function, one might consider that a value is a number, not a pair of numbers or a point in the plane. Any answer that is not a number is essentially wrong. 9. Textbook,.8, Exercise # 48, p. 949: Find the absolute maximum and the absolute minimum values of the function f(x, y) = x + y x y on the closed triangle of vertices (, ), (, ) and (, ). Notice that it is the region bounded by the lines x =, y = and y = x. The boundary can be thus described as the union of three segments that I will label by I, II, and III, and they are I) x =, y, II) x, y =, III x, y = x. We begin finding critical points, setting the first order partial derivatives to : f x = x =, f y = y =. The only critical point is (, ); it is on the boundary of the region. It will probably pop up again when studying the function on the boundary. Boundary Behavior. On I. Let h(y) = f(, y) = y y. Points where something could happen are the endpoints of the interval or at a critical point of h. Now h (y) = y = for y =. So the points we have to consider here are the points (, ), (, ), (, ). On II. Since the function is symmetric in x and y, the points to consider on this interval will be (, ), (, ) and (, ). On III. Let h(x) = f(x, x) = x 4x. Now h (x) = 4x 4 = for x = so a point to consider is (, ). We also have to consider the endpoints, but these are (, ) and (, ); already considered before. All in all we have the following points to consider: We have: (, ), (, ), (, ), (, ), (, ), (, ). f(, ) =, f(, ) =, f(, ) =, f(, ) =, f(, ) =, f(, ) =. From this we see that the minimum value is (assumed at (, )), the maximum value is (assumed at (, ), (, ), and at (, )).. Find the maximum and the minimum value of the following function in the indicated domain. If there is no maximum or minimum, say so. f(x, y) = 3 x y in the closed disc {(x, y) : x + y 9}.

8 8 This is a continuous function in a closed and bounded domain. It MUST assume a maximum value and it MUST assume a minimum value. f(x, y) is negative, but it is for x + y = 3. The minimum value is, assumed one the circle of radius 3 centered at the origin. Moving away from that circle, the function grows. It has a local maximum at (, ), with a value of 3. However on the boundary of the domain the value assumed is 6. The answer is: The minimum value is, the maximum value is 6.. ompute xy da; is the region in the first quadrant bounded by x =, y = x and y = 8 x. (x + y) da = 8 x x (x + y) dy dx = Write as an iterated integral in the order dx dy the integral of f over the region in quadrants and 3 bounded by the semicircle of radius 3 centered at (, ) y f(x, y) dx dy. 3. Evaluate y da where is the region bounded by y =, y = x, y = x. y da = y +y y dx dy =. 4. Find the volume of the solid in the first octant bounded by the coordinate planes and the surface z = y x. The solid can be described as being {(x, y, z) : x, y x, z y x }. Thus V = x 5. Sketch the region of integration and evaluate the ( y x ) dy dx = x xe y dy dx by reversing the order of integration. 4 y The region can be described by = {(x, y) : x, y 4 x }. Here s a sketch. From the sketch, or directly, we see that the region can also be described as = { y 4, x 4 y}. Thus 4 x xe y dy dx = 4 y 4 4 y xe y 4 y dx dy = 4 ( 4 y) ey 4 y dy = 4 e y dy = e8. 4

9 9 6. Find the volume of the solid bounded by the paraboloids z = x + y and z = 7 x y. The paraboloids intersect for x + y = 9 so the solid can be described by D = {(x, y, z) : x + y 9, x + y z 7 x y }. The description in terms of polar coordinates is D = { r 3, θ π, r ( + cos θ) z 7 r ( + sin θ)}. Thus V = π 3 [ (7 r ( + sin θ)) r ( + cos θ) ] r dr dθ = π 3 (7 3r )r dr dθ = 43 π. 7. Sketch the region inside both the cardioid r = cos θ and the circle r =, and find its area. Sketch: The region is symmetric with respect to the x axis, so the area is twice the area above the x-axis; that is for θ π. For θ π/, r goes from the origin to the cardioid; after that to the circle. Thus ( π/ cos θ π ) ( ) π/ A = r dr dθ + r dr dθ = ( cos θ) dθ + π = 5π π/ Find the average distance of points within the cardioid r = + cos θ and the origin. The distance of a point from the origin is r, so the average will be the integral of r over the cardioid, divided by the area of the cardioid. The area of the cardioid is The average is 9. ompute In polar coordinates Av = A A = π +cos θ π +cos θ x y x + y da, is the unit disc. + x y x + y + da = r dr dθ = 3π. r dr dθ = A 5π 3 = 9. π Since π cos θ dθ = π sin θ dθ =, the integral works out to. 3. The figure shows the region of integration for the integral x x r(cosθ sin θ) + r r dθ dr. f(x, y, z) dz dy dx.

10 ewrite the integral as an iterated integral in the five other orders. That is, replace the question marks with appropriate expressions so the following five integrals are equal to the one given above.?????? f(x, y, z) dy dz dx,??? The region can be described by???????? f(x, y, z) dx dy dz,? f(x, y, z) dy dx dz,??????????? f(x, y, z) dx dz dy. D = {(x, y, z) : x, y x, z x }.? f(x, y, z) dz, dx dy, I usually prefer to work from the description. If there is a picture I ll use it as a guide, but pictures are not essential. Here are the other five integrals, in the same order as in the question. Actually, the last two are a bit tricky and will result in more than one integral. You will see. Because of this, see also the grading note at the end. dy dz dx) There is no relation between y and z in the description, so this is quite easy, x x f(x, y, z) dz dy dx = x x f(x, y, z) dy dz dx. dy dx dz) z can range from to ; since z x we get x z. For every value of (x, z), y still can go from to x. Thus x x dz dx dy) There isn t much of a problem here. x x f(x, y, z) dz dy dx = f(x, y, z) dz dy dx = z x y x f(x, y, z) dy dx dz. f(x, y, z) dz dx dy. dx dy dz) This one is tricky. We have z and y. But when it comes to x we have to have both x y and x z. For a given value of z, z, where x ranges depends on whether y z; i.e., y z, in which case x goes from to y, or if y > z, in which case x ranges from to z. The answer in this case is x x f(x, y, z) dz dy dx = z z + z y f(x, y, z) dx dy dz f(x, y, z) dx dy dz.

11 dx dz dy) The analysis here is similar as in the previous case. The answer works out to x x f(x, y, z) dz dy dx = 3. Suppose f is continuous and a >. Show that a y z ( y) y + f(x) dx dz dy = a ( y) z (a x) f(x) dx. f(x, y, z) dx dz dy f(x, y, z) dx dz dy. The region of integration is D = {(x, y, z) : y a, z y, x z}. It can also be described by D = {(x, y, z) : x a, x z a, z y a}. Thus a y z f(x) dx dz dy = = a a a a x f(x) z a x f(x) dy dz dx = a (a z) dz dx = f(x) a a a x z dy dz dx (a x) f(x) dx. An alternative way of doing this exercise is describing the region in the form D = {(x, y, z) : x a, x z a, x y z}. In this case a y z f(x) dx dz dy = = a a y a x f(x) x a x f(x) dz dy dx = a (x y) dy dx = 3. Let = {(x, y) : x π, y π}. Evaluate sin ( max(x, y ) ) da. f(x) a By max(a, b), if a, b are numbers, one understands a if a b; otherwise it is b. a y x x dz dy dx (a x) f(x) dx. We divide the square up into the triangle T where x < y, and the triangle T where x > y: T = {(x, y) : y π, x y}, Then sin ( max(x, y ) ) da = = sin y da + T sin x da = T π 33. Use cylindrical coordinates to evaluate 3 y sin y dy + 3 π 9 x T = {(x, y) : x π, y x}. π y sin y dx dy + x sin x dy = cos(y ) + x dz dy dx. + y π π x ( + cos(x ) sin x dy dx The z integral is totally independent from the x, y integrals, and {(x, y) : 3 x 3, y 9 x } is the half circle of radius 3 above the x-axis, centered at the origin. 3 9 x 3 + x + y dz dy dx = 3 ( 3 = 3 9 x r + r dr + x dy dx = + y ) ( π ) dθ = π ln. 3 π π r dθ dr + r ) =.

12 34. Let D be the region in the first octant of 3 bounded by the paraboloid z = 4 (x +y ). ompute x dv. D The picture on the right shows the paraboloid and the plane z =. Hint: Use cylindrical coordinates. D x dv = π/ 4 r r cos θ dz dθ dr = π/ (4 r )r cos θ dθ dr = (4r r 4 ) dr = Let D = {(x, y, z) : x, y, z, x + y + z 9}. ompute e z dv. using spherical coordinates. Working with spherical coordinates, 3 π/ π/ e z dv = e ρ cos ϕ ρ sin ϕ dθ dϕ dρ = π 3 π/ e ρ cos ϕ ρ sin ϕ dϕ dρ D = π ( 3 ρ ) ϕ=π/ ρ eρ cos ϕ dρ = π 3 ( ρ(e ρ ) dρ = π e 3 7 ). 4 D ϕ= 36. Find the volume of the region common to the three cylinders x + y, y + z, z + x. There is a picture of the region on page 44 of the textbook. This is not as hard as it seems, one merely has to be systematic. In the first place, we can simplify the problem by using some symmetry, finding out the volume of the intersection in the first octant x ge, y, z ; the total volume is then 8 times the volume in the first octant. If (x, y, z) are in the intersection of the three cylinders then they verify simultaneously x + y, y + z, z + x. So, since we are in the first octant, beginning with x, we have that we can have any value of x in the interval [, ], so x. Next we see that since x + y, we ll have y x. The problem is now what happens with z. The answer is that since we must have both z x and z y we need to break up the unit disc (more precisely the quarter unit disc) into the part where x y and where x y.a simple ollege Algebra (i.e., High School) level computation shows that x y if and only if y x. We could continue with cartesian coordinates, but here is where switching t cylindrical coordinates makes life easier. In the quarter disc below, the part in red is where y x, so there x y and z will range from to x = r cos θ. In the green part, y x, so there z will range from to y = r sin θ.

13 3 In polar coordinates, the red region is described by {(r, θ), : r, the π/4} while the green part is {(r, θ), : r, π/4 the π/}. Putting this all together, remembering to multiply by 8, we see that ( π/4 π/ ) V = 8 r cos θ r dr dθ + r sin θ r dr dθ. The integrals seem hard, but they are not really. In the first iterated integral, we can perform the integration with respect to r by the substitution s = r cos θ, so rdr = /( cos θ) ds and π/4 r cos θ r dr dθ = = 3 = 3 = 3 π/4 π/4 π/4 π/4 s ds = cos θ sin θ 3 π/4 sin 3 θ cos dθ = θ 3 ( cos θ sin θ cos θ + sin θ ( ( ) ( ) ) π/4 cos θ s3/ sin θ dθ sin θ( cos θ) cos dθ θ ) dθ = π/4 3 (tan θ sec θ cos θ) =. Well, maybe it got a bit hairy at the end, but there is always Wolfram alpha to compute integrals (or Maple, or Mathematica, or tables). A similar computations shows that one also has π/ π/4 so that finally one gets V = 8( ). r sin θ r dr dθ =, 37. The result of this exercise will be used in Exercise 38. Find the coordinates of the center of mass (centroid) of the portion of the cardioid r = + cos θ above the x-axis; that is, the region described by = {(r, θ) : θ π, r + cos θ}.

14 4 We get thus A = M y = M x = π +cos θ π +cos θ π +cos θ x = M y A = 5 6, r dr dθ = 3π 4, r cos θ dr dθ = 5π 8, r sin θ dr dθ = 4 3. ȳ = M x A = 6 9π. 38. We are going to work here with the solid obtained by revolving a cardioid. We flip the cardioid of Exercise 37 so it looks like: We then rotate it about the z-axis. (a) Show, or at least convince yourself, that in spherical coordinates the solid so obtained is described by (b) Use spherical coordinates to find its volume. D = {(ρ, ϕ, θ) : θ π, ϕ π, ρ + cos ϕ}. (c) A theorem due to Pappus (who lived some,7 years ago) states that the volume of a solid obtained by rotating a plane figure of area A about an axis equals A, where is the circumference of the circle described by the centroid of the figure. The solid of this problem is obtained by revolving about the z-axis the portion of the cardioid described in Exercise 37, except that what was the x-axis in Exercise 37 is now the z-axis. If the coordinates of the center of mass you found in Exercise 37 are (a, b), the center of mass of the figure rotated about the z-axis are at (b,, a). Think about it. Use this information to verify Pappus Theorem in this case.

15 5 (a) onvince yourself! (b) By the description in spherical coordinates, V = π π +cos ϕ ρ sin ϕ dρ dϕ dθ = π 3 π ( + cos ϕ) 3 = 8π 3. (c) A bit of reflection (no pun intended) shows that the cardioid to be rotated comes from the one in the previous exercise if the xaxis is replaced by the z axis and is placed in the plane y = : The coordinates of the center of mass of the cardioid in this position are ( 6 9π,, 5 6 ); in particular, the distance of the center of mass from axis of rotation is 6 9π. This means that the center of mass will describe a circle of radius 6/9π. By Pappus theorem (recalling that the area was 3π/4), V = π 6 9π 3π 4 = 8π A solid is bounded above by the paraboloid z = 4 x y and below by the cone z = x + y. The density of the solid satisfies ρ(x, y, z) = z. Find the coordinates of the center of mass. Note: This exercise results into some really nasty integrals. While you will not have access (at least not legally) to any computer algebra for the final, you might consider using something like Wolfram alpha here. In cylindrical coordinates the two surfaces bounding the region are z = 4 r and z = r. They come together for r = ( 7 )/. The region can be described as 7 D = {(r, θ, z) : r, θ π, r z 4 r }. By symmetry, x = ȳ =. We have (integrals computed by Maple) V = M xy = Finally, after some simplification ( 7 )/ π 4 r r ( 7 )/ π 4 r z = r z r dz dθ dr = π, 4 z r dz dθ dr = π (Here finding an approximate value was a good idea, to see if the answer is at least likely. It is.) 4. Let be the region in the first quadrant bounded by the curves xy =, xy = 5, y = 3x, y = 3x +.

16 6 Show that changing variables by u = xy, v = y 3x, changes the region into a rectangle {(u, v) : a u b, c v d}. Find the correct values of c and d and use the change of variables to compute (y + 3x) da. ompute also the integral without changing variables and verify that you get the same result. ( ) (x, y) (u, v) Here is a fact that could ease computations: (u, v) =. (x, y) It is easy to see that the indicated change maps the region into the rectangle { u 5, v }. The only place where things might get slightly hairy is when trying to invert the map, get x, y in terms of u, v. From u = xy we have y = u/x; substituting into the expression for v we get 3x + vx u =, from which x = v ± v + u. 6 v + u v learly x has to be positive so x =. Then 6 y = u x = 6u v + u = 3( v + u v). We now have to compute the Jacobian determinant and here is where things get really nasty, having to compute (x, y) (u, v) = x y u v x y v u. But perhaps the hint can help. We have Usually this would be of some help; to get (u, v) (x, y) = u v x y u v = y x( 3) = y + 3x. y x (x, y) (u, v) (x, y) (u, v) = (u,v) (x,y) from this we have to use that = 3x + y, and replace x, y by their expressions in u, v. We also have to replace any other x, y in the integral by their expressions in u, v. But in this case the simplification is absolute, 5 (y + 3x) da = (y(u, v) + 3y(u, v)) (x, y) (u, v) dv du = 5 (y(u, v) + 3y(u, v)) dv du = y(u, v) + 3y(u, v) 5 And solving for x, y was unnecessary! (But that s because the integral was so set up.) If we find the integral without changing we get dv du = 6. (y + 3x) da = same as before, of course. 3x+ (y + 3x) dy dx + 5/3 5/x /3 /x 3x (y + 3x) dy dx = 6,

17 7 4. Evaluate xy da where is the region in the plane bounded by the curves y = x /5, y = (x ) /5+, y = x and y = x 4 (see picture below) by changing variables by x = u + 5v, y = u + v. In this, as in all change of variables exercises, you must clearly indicate what the transformation does; into what it changes the region of integration, and give at least a good reason why the new region is what you say it is. omputing the integral by any method except the indicated one does not count!. Let = {(u, v) : u, v }. The boundary of consists of the 4 segments I = {(u, ) : u }, I = {(, v) : v }, I 3 = {(u, ) : u }, I 4 = {(, v) : v }. The change carries I to points satisfying x = u, y = u, u. This is a parametrization of the segment y = x from (, ) to (, ), the left boundary of. The segment I gets mapped to points satisfying x = + 5v, y = + v, v. This is a parametrization of the arc of parabola y = + (x ) /5, from (, ) to (6, ), the upper boundary of. The segment I 3 maps to x = u + 5, y = u +, u, a parametrization of the line segment from (5, ) to (6, ), the right boundary of. Finally the segment I 4 maps to the curve parameterized by x = 5v, y = v, v, the arc of parabola of equation y = x /5 from (, ) to (5, ), the lower boundary of. When a smooth map maps the boundary of a simple region to the boundary of a simple region. it will map the interior either to the interior or to the exterior of the second region. It is easy to see that points inside stay inside ; it suffice to check at a single point. Let us compute the Jacobian of the change: (x, y) (u, v) = 5 v = v 5 Incidentally, the fact that the Jacobian is negative (v 5 < if v ) indicates that the mapping is orientation reversing. You can notice that if you go through the boundary of with the interior to your left, you traverse the boundary of with the interior to your right. The rest is easy, I think: xy da = 4. D is the region bounded by the planes (u + 5v)(u + v ) v 5 du dv = (u + 5v)(u + v )(5 v) du dv = x + y + 3z =, x + y + 3z =, 4x y =, 4x y =, z = 3, z =. Use a change of variables to compute D (5x + y) dv.

18 8 An obvious choice is u = x + y + 3z, v = 4x y, w = z This changes the region of integration to u, v, 3 w. We need to rewrite 5x + y in (x, y, z) terms of u, v, w and calculate the Jacobian determinant. For both it may seem necessary to express (u, v, w) x, y, z in terms of u and v. In general, this may be necessary, but there are shortcuts. For example we see (x, y, z) easily that u + v = 5x + y + 3z so 5x + y = u + v 3w. To compute we can use the following rule (u, v, w) (x, y, z) (u, v, w) = (u, v, w) (x, y, z), but then replace in the last expression every appearance of x, y, z by its expression in terms of u, v, w. If we proceed this way we have (u, v, w) (x, y, z) = 3 4 = 9 Since no x, y, z appear, we now can say that D (5x + y) dv = (x, y, z) = /9. We thus have (u, v, w) 3 (u + v 3w) 9 du dv du = 3. A more systematic approach has one solving x, y, z in terms of u, v, w. This works out to Thus (as before) and and the rest is as before. x = 9 u + 9 v 3 w, y = 4 9 u 9 v 4 3 w, z = w (x, y, z) (u, v, w) = = 9 ( 5x + y = 5 9 u + 9 v ) 3 w u 9 v 4 3 w = u + v 3w 43. The picture shows a number of level curves of the function z = x +y or, as is equivalent, equipotential curves of the potential V (x, y) = x + y

19 9 (a) Show that at every point (x, y) the equipotential curve through that point has a tangent perpendicular to the gradient. (b) Illustrate this by selecting several points (at least 3, no more than 5) and drawing at those points the gradient and the tangent to the curve. (a) The level curves have equations x + y =. By implicit differentiation y = x. So at every point (x, y) the slope of the tangent line of the level curve through that point is x. We need a vector with that slope; the easiest one is probably, x. At every point (x, y), the vector, x is tangent to the level curve through that point. On the other hand the gradient of V at (x, y) is V (x, y) = x,. Since x,, x = x x =, we see that at every point (x, y) the equipotential curve through that point has a tangent perpendicular to the gradient. (b) In the picture below 5 points were marked; the tangent vectors are in red, the gradients in blue.

20 44. A particle of mass m is moving under the action of a conservative force field F = V. Let r(t) = x(t)i + y(t)j + z(t)k be the position of the particle at time t and v(t) = r (t) its speed. Show that the energy E(t) = mv(t) + V (x(t), y(t), z(t)) remains constant. Notice that so that, v(t) = r (t) r (t) d ( v(t) ) = r (t) r (t) + r (t) r (t) = r (t) ɛ (t) dt and, taking also into consideration Newton s nd law: mr = F, and the chain law, de dt (t) = m d ( v(t) ) + d V (x(t), y(t), z(t)) dt dt = mr (t) r (t) + V x (x(t), y(t), z(t))x (t) + V y (x(t), y(t), z(t))y (t) + V z (x(t), y(t), z(t))z (t) = F(x(t), y(t), z(t)) r (t) + V (x(t), y(t), z(t)) r (t) = (F(x(t), y(t), z(t)) + V (x(t), y(t), z(t))) r (t) = since F(x(t), y(t), z(t)) + V (x(t), y(t), z(t)) = V (x(t), y(t), z(t)) + V (x(t), y(t), z(t)) =. Since the derivative of E is identically, we conclude E is constant. 45. Escape Velocity. We compute the escape velocity, or rather the escape speed, from a planet of radius, surface acceleration of gravity g. In this model, only the planet exists in the whole universe. This model is not too bad as long as we don t get too far away from the planet. Infinity in physics may just be a large number. We assume a rocket is fired from the surface of the planet, in a direction perpendicular to the surface, with an initial speed v. The question to be answered is: What should v be so that the rocket does not return to the planet? We assume the rocket is so small that its gravitational pull on the planet is negligible. The gravitational force field created by the planet is then, according to Newton s law of universal gravitation: where Here are your instructions: m = mass of the rocket, M = mass of the planet, F = GmM r 3 r, r = the radius vector with origin at the center of the planet, G = the gravitational constant. (a) Because G, M can be hard to compute, while g, are not, get rid of G, M by using the fact that on the planet surface F = mg; using this show that GM = g. (b) Show that F is conservative and find V such that F = V. V is only determined up to a constant; a convenient constant is one such that lim x +y +z V (x, y, z) =. Using this constant will result in a negative V. (c) As seen in Exercise 44, the energy E(t) = V (r(t)) + m r (t) is constant. As the speed of the rocket decreases, V increases accordingly (becoming less negative). Once the speed of the rocket is, the rocket will start returning to the planet. The idea now is to figure out what v should be so that this only happens at infinity, where V is equal to. Equating the energy with the limit for r and the value when r = should produce the escape speed.

21 (d) For the earth g 9.8m.s and 6, 4, m. alculate the escape speed from earth. Most of everything is immediate. Part of the exercise (seeing the gravitational field is conservative and calculating the potential was done in class). (a) On the surface of the planet we have r = so that the intensity of the force is given both by F = GmM and by mg. Equating one gets MG = g (b) Writing F = F, F, F 3, r = x, y, z, we have and mg x F = (x + y + z ), F mg y 3/ = (x + y + z ), F mg z 3/ 3 = (x + y + z ) 3/ F y = 3mg xy (x + y + z ) = F 5/ x, F z = 3mg xz (x + y + z ) = F 3 5/ x, F y = 3mg yz (x + y + z ) = F 3 5/ y. The field is conservative. We want V = F so we begin with V = F dx = mg x(x + y + z ) 3/ dx = mg (x + y + z ) / + h(y, z). Proceeding as usual one gets as next step that h y =, so h(z) = g(z). Finally one also gets g (z) =. So we can take V (x, y, z) = mg (x + y + z ) / ; this is in fact the determination that satisfies lim r V =. (c) For the rocket we will have E(t) = V (x(t), y(t), z(t)) + m v(t) is constant. At time t = (launching time) we have v() = v while the potential energy is mg / = mg. The total energy, which will remain constant during the total life of the rocket, is E = mg + v. We want the kinetic energy given to the rocket to be sufficient to take it to infinity, so it shouldn t be exhausted before then. At infinity the potential energy is maximum (equal to, which is maximum since it is otherwise negative) so the total energy, if the rocket s velocity is zero at infinity (it just got there!) will also be zero, so E =. onstancy of energy now implies mg + mv =. Thus v = g. (d) Taking the given values of g, one gets v, m/s, or approximately km/s. 46. Evaluate the following line integrals over the given curves (a) ( points) xe yz ds where is the curve parameterized by r(t) = t, t, 4t, t. xe yz ds = te 8t = te 8t dt = 6 ( e 8 ). (b) ( points) x ds, where is given by r(t) = t, 4t, t, t =,. It may help to know that x a + bx dx = 3b (a + bx ) 3/ +. x ds = t t dt = t 7 + 4t dt = (333/ 7 3/ ).

22 (c) (x y + z) ds, where is a circle of radius 3 in the plane x =, centered at (,, ). (x y + z) ds = = 3 π π ( 3 cos t + 6 sin t) + ( 3 sin t) + (3 cos t) dt ( 3 cos t + 6 sin t) dt = 6π. xy (d) ds where is the line segment from (, 4, ) to (3, 6, 3), followed by the arc of circle centered at z (,, 3) in the plane z = 3 from (3, 6, 3) to (6, 3, 3); clockwise. Finding a parametrization for the arc of circle is slightly complicated. It is an arc of a circle of radius = 45. If we were going counterclockwise we would go from θ = arctan / to θ = arctan. But the orientation plays no role in this type of line integral, so we may use the counterclockwise parameterization. (e) Evaluate xy z ds = = 8 = 5. ( + t)(4 + t) arctan + + t + dt + (4 + t) dt arctan arctan / ( 45 cos t) 45 sin t 45 dt arctan / 3 cos t sin t dt = 4 + arctan 45 5 sin t z ds where is the curve r(t) = cos t sin t, cos t + sin t, cos t, t π. z ds = π = ( cos t) ( sin t cos t) + ( sin t + cos t) + ( sin t) dt π cos(t) + 4 sin t dt =. 47. Find the work done by the force field F = y, x, z in moving a particle along the helix arctan / for t π. x = cos t, y = sin t, z = t π W = = helix π y, x, z dr = π ( 4 + t ) 4π dt = 8π +. cos t, sin t, t π sin t, cos t, π dt 48. (a) For what values of a, b, c and d is the field F = ax + by, cx + dy conservative? so the answer is b = c. (ax + by) y = b, (cx + dy) x = c

23 3 (b) For what values of a, b, and c is the field F = ax by, cxy conservative? so the answer is c = b. 49. onsider the vector field (ax by ) y = by, (cxy) x F(x, y, z) = yz +, xz + y, xy + 3z. = cy (a) Show that F is conservative and determine a function f so that F = f. (b) Let be the curve given in vector notation by Evaluate r(t) = t i + t 3 j + k, t. + t F dr. ealizing that these questions are related may save some time. (a) We have y (yz + ) = z (xz + y), x z (yz + ) = y = x (xy + 3z ), z (xz + y) = x = y (xy + 3z ). Since the functions are nicely differentiable everywhere, the fact that the cross derivatives are equal is not only necessary, but also sufficient for the field to be conservative. We proceed to find f so f = F. To satisfy f x = yz + we integrate with respect to x. f(x, y, z) = (yz + ) dx = xyz + x + h(y, z). To get f y = xz + y we need so that y h (xyz + x + h(y, z)) = xz + y, hence xz + (y, z) = xz + y, y thus h h(y, z) = (y) dy = y + g(z). (x, z) = y y At this point, f(x, y, z) = xyz + x + y + g(z); to get the final condition, namely, ϕ (x, y, z) = xy + 3z z we need (and it suffices) to have ( xyz + x + y + g(z) ) = xy + 3z, z integrating with respect to z we see that we can take g(z) = z 3. Thus f(x, y, z) = xyz + x + y + z 3. (b) The path starts at (,, ) and ends at (,, ). Since the vector field is conservative, F dr = f(,, ) f(,, ) = 4 8 = 4.

24 4 5. (a) Show that the vector field F = xe y + z e x, x e y + ye z, y e z + ze x + is conservative and find the function f such that f = F. We have y (xey + z e x ) = xe y = x (x e y + ye z ), z (xey + z e x ) = ze x = x (y e z + ze x + ), z (x e y + ye z ) = ye z = y (y e z + ze x + ). The field is conservative. Now f(x, y, z) = (xe y + z e x ) dx = x e y + z e x + h(y, z). The function h has to satisfy so that ( x e y + z e x + h(y, z) ) = x e y + ye z ; i.e. x e y + h y y (y, z) = x e + ye z h y (y, z) = yez, hence h(y, z) = y e z + g(z). eturning with this to the expression for f we have so far f(x, y, z) = x e y + z e x + y e z + g(z). We still need to satisfy f z (x, y, z) = y e z + ze x + or z (x e y + z e x + y e z + g(z)) = y e z + ze x + ; i.e.ze x + y e z + g (z) = y e z + ze x +. This forces g (z) = ; we take g(z) = z and we get f(x, y, z) = x e y + z e x + y e z + z. (b) An object is moved along the path parameterized by x = t, t, z = t 3 from time t = to time t =. Find the work done by the force field of part (a). The particle is at (,, ) for t = and at (, 4, 8) for t =. Since the field is conservative, the work done is f(, 4, 8) f(,, ) = 4e e + 6e Evaluate xy dx + xy dy in two ways where is the positively (counterclockwise) oriented triangle of vertices (, ), (, ), and (, ).: (a) Directly, using the definition. (b) Using Green s Theorem. (a) The triangle is the sum of three line segments that I will denote by I, II, and III, where the parameterizations are I. x(t) = t, y(t) = t, t (Segment from (, ) to (, ).) II. x(t) = t, y(t) =, t (Segment from (, ) to (, ).) III. x(t) =, y(t) = t, t (Segment from (, ) to (, ).)

25 5 We now have Thus xy dx + xy dy = I I II III xy dx + xy dy = xy dx + xy dy = xy dx + xy dy = (t 3 + t ) dt = 7, ( t) dt =, xy dx + xy dy + xy dx + xy dy + (y dx + xy dy = 7 II III =. (b) By Green s Theorem, since the triangle can be described by {(x, y) : y, x y}, we have y ( ) xy y xy dx + xy dy = x xy dy dx = (y xy) dy dx = (y y 3 ) dy = y. 5. Let be the triangle in the (x, y)-plane of vertices (, ), (4, ), and (8, 8). You are supposed to calculate x da in four different ways, which are parts (a),(b),(c), and (d) of this exercise. (a) alculate x da if is the triangle of vertices (, ), (4, ), (8, 8) directly, as an iterated integral (or a sum of iterated integrals). As a help, the right boundary of the triangle lies on the line y = 3 x 4, the lower boundary on the line y = x/. The region can be described as the union of {(x, y) : x 4, x y x} and {(x, y) : 4 x 8, 3 x 4 y x}. Thus 4 x da = x x/ x dy dx + 8 x 4 x dy dx = 3 x 4 4 x dx x(4 x) dx = 3. (b) alculate x da if is the triangle of vertices (, ), (4, ), (8, 8) by Green s Theorem, using the fact that x = ( ) x x. By Green s Theorem ( ) x x da = x dy where is the positively oriented boundary of the triangle. The boundary is made up of the segments L, followed by L, followed by L 3 where L joins (, ) to (4, ), L joins (4, ) to (8, 8) and L 3 joins (8, 8) to (, ). the corresponding parameterizations are We have L L x dy = x dy = L 3 x dy = L : x(t) = 4t, y(t) = t, t, L : x(t) = 4 + 4t, y(t) = + 6t, t, L 3 : x(t) = 8( t), y(t) = 8( t), t. (4t) () dt = 6 (4 + 4t) (6) dt = 48 t dt = 6 3. [8( t)] ( 8) dt = 56 ( + t) dt =. ( t) dt = 56 3.

26 6 Thus (c) alculate fact that x = (xy) y. By Green s Theorem x da = L x dy + L x dy + L 3 x dy = = 3. x da if is the triangle of vertices (, ), (4, ), (8, 8) by Green s Theorem, using the (xy) y da = dx where is the positively oriented boundary of the triangle. We use the same parameterizations of the previous point, getting Thus (d) alculate xy dx = L xy dx = L xy dx = L 3 8t (4) dt = 3 t dt = 3 3. (4 + 4t)( + 6t)(4) dt = 3 [8( t)] ( 8) dt = 5 x da = xy dy xy dy xy dy = 3 L L L 3 3 (3t + 4t + ) dt = 8. ( t) dt = = 3. x da if is the triangle of vertices (, ), (4, ), (8, 8) by changing variables by u = 3x y v = x + y In this, as in all change of variables exercises, you must clearly indicate what the transformation does; into what it changes the region of integration, and give at least a good reason why the new region is what you say it is. We see that the transformation keeps (, ) and (8, 8) fixed and carries (4, ) into (8, ). Being linear, it will carry lines to lines, so the sides go to the sides of the triangle, and so does the interior. Inverting the transformation, we get x = u + v y = 4 u v. The Jacobian of the transformation is (x, y) (u, v) = / / /4 3/4 = 4. Thus x da = 8 u ( (u + v) 4 dv du = 8 8 u 53. Use Green s Theorem to evaluate the following line integrals. (u + v) dv du = u du = 3.

27 (a) (b) ( 3y + x 3/ ) dx + (x y /3 ) dy where is the boundary of the halfdisc {(x, y) : x + y, y } with positive (counterclockwise) orientation. Let = {(x, y) : x + y, y }. By Green s Theorem, ( 3y + x 3/ ) dx + (x y /3 ) dy = ( ( 3)) da = 4 da4π. (x 3 y ) dx (3x y + xy) dy, where is the square of vertices (±, ±), with positive (counterclockwise) orientation. By Green s Theorem (x 3 y ) dx (3x y + xy) dy = ( 6xy + y) da =. 54. ( points) Use Green s Theorem to find the area of the region bounded by the counterclockwise circular arc from (, ) to (, ) of radius, centered at (, ), and the clockwise semicircle of radius / centered at (/, /). See picture. 7 heck your answer computing the area using elementary school geometry. The large circular arc can be parameterized by x = cos t, y = sin t, t π while the counterclockwise version of the lower arc can be parameterized by x = + cos t, + sin t, π/4 t 5π/4. Then, using the version A = x dy, A = = 3π/ 3π/ (cos t)(cos t) dt cos t dt 5π/4 π/4 5π/4 π/4 ( + cos t)( cos t) dt cos t dt 5π/4 π/4 cos t dt = 3π + π 4 = π +. An elementary school approach could be as follows. The region can be obtained by first cutting of from the unit circle a circular segment spanning an inner angle of 9 : What remains is three quarters of the circle and a right triangle of legs of length. By very elementary geometry, the area is 3 4 π +. To get the figure of the problem we have to cut out next a semicircle of radius /, which has area ( ) π = π 4. Thus the area of the region is 3π 4 + π 4 = π Find the area of the shaded figure in two ways: (a) Using Green s Theorem. (b) Not using Green s Theorem, for example using double integrals and or elementary geometry

28 8 The boundary of the figure consists of 4 one quarter circle arcs as shown in the picture below ed: Quarter circle of radius centered at (, ) Blue: Quarter circle of radius centered at (3, ) Green: Quarter circle of radius centered at (3, ) Yellow: Quarter circle of radius centered at (, ) By Green s Theorem Denote the arcs by (red), (blue), 3 (green), and 4 yellow. It will probably be easiest to parameterize them as arcs of a circle gone through counterclockwise, then subtract the integral if this direction is not the one we want. If this is unclear, I hope to clear it up in the execution.. We parameterize by x = cos t, y = + sin t, π t.

29 9 Thus x dy =. The easy thing is to parameterize by π/ ( cos t)( cos t) dt = 4 x = 3 + cos t, y = + sin t, π/ cos t dt = π. π t π. The problem is that then we are going through the arc in the wrong direction. ather than figure out how to parameterize so as to get the right direction of traversal, we simply subtract the integral rather than add it. π π x dy = (3 + cos t)(cos t) dt = (3 cos t + cos t) dt = 3 + π π/ π/ We parameterize by Thus x dy = 4. We parameterize by x = 3 + cos t, y = + sin t, π/ (3 + cos t)( cos t) dt = π/ π t π. x = cos t, y = + sin t, π t, but this is again the wrong direction. So 4 will have to be subtracted. 4 x dy = π/ (cos t)(cos t) dt = (6 cos t + 4 cos t dt = 6 + π. π/ cos t dt = π 4. We thus have A = x dy = x dy x dy + x dy x dy = π ( 3 + π ) 6 + π π 4 = 3π 3. Not by Green s Theorem. We see that the region is what remains of the 3 3 square after removal of the brown an yellow regions.

30 3 These regions are symmetric. The brown region is divided into 3 regions I labeled S,, D. The area of the region we are considering is equal to A = 9 (S + + D), where I am using the same letters to denote the regions that make up the brown region as their areas. Now, is a rectangle, so =. D is a quarter circle of radius, so D = π/4. The region S extends from x = to x = and it is not too hard to figure out the equation of the curve bounding it at the top, namely. It is part of the circle x + (y ) = 4, so we see it is y = 4 x. Thus S = 4 x dy dx = ( 4 x ) dx = 4 4 x dx. We can look this last integral up in tables. Or we can substitute x = sin t. Or we can realize it is the area of a quarter circle of radius, thus equal to π. Thus S = 4 π. Finally, the area A we are looking for works out to A = 9 (4 π + + π 4 ) = 3π 3 as before. 56. The cycloid is a curve apparently first mentioned by Galileo. It became famous thanks to the Bernoulli brothers discovering that it was both the brachistochrone and the tautochrone (look it up!). Possibly the arches of the Ponte Vecchio in Florence Italy have a cycloidal shape. Ponte Vecchio over the Arno A parametrization of a single arc of the cycloid is given by x = a(t sin t), y = a( cos t), t π, where a > is a constant. Use Green s Theorem to find the area of the region bounded by the arc of a cycloid and the x axis. (The first person to compute this area, as well as finding the length of an arc of the cycloid, was Gilles Personne de oberval (6-675); they were also computed only a little bit later, independently, by Evangelista Torricelli (68-647), more famous for having invented the barometer.) We ll use the formula A = (x dy y dx) where is the boundary of the region. Notice that this integral is zero on the bottom part of the boundary; the segment from (, ) to (π, ) because dy =, x = on that segment. However, one thing that we have to careful with is the orientation. I ll do the calculations and we ll see why. A = = a π π = 3πa. (a(t sin t)(a sin t) a( cos t)a( cos t)) dt ( t sin t sin t + cos t cos t ) dt

31 3 But this can t be, area cannot be negative!!! Something must be wrong, and the wrong isn t just righted by merely changing the sign for no good reason. A bit of thinking shows where we went wrong. The parametrization has us go through the arc of the cycloid starting at (, ) and ending at (πa, ). Doing this, the region bounded by this arc is to our right. That means we are going through the boundary in a negative orientation. That means we get the negative of what we would get by going through in a positive orientation. Now we can honestly change the sign and say A = 3πa. 57. Find the area of the region bounded by the curves y = sin x, x = π + sin y, y = π and x =. We ll use the formula A = where is the boundary of the region. The boundary is made up of 4 parts I ll denote by A, B,, D, as pictured. x dy, The very unskillfully drawn arrowheads indicate the positive orientation. The A section can be parameterized by x = t, y = sin t, t π, to give π x dy = t sin t dt =. A Or, we could have just said we use x as a parameter and written it directly as For B we can use y as a parameter; the range is again π, so that π x dy = (π + sin y) dy = π +. B π x d(sin x) = π x cos x dx. And we are done because on we have dy =, on D we have x =, so that the integrals over and D are zero. Adding up we see the area is π.