MAC 232-593-Calculus II Spring 23 Homework #2 SOLUTIONS Note. Beginning already with this homework, sloppiness costs points. Missing dx (or du, depending on the variable of integration), missing parentheses, incorrect use of the equal sign; all will be subject to points being taken away. The exact number of points will vary so, to be on the safe side, be careful. As one example of how things should be, if I see an integral sign ( ), I should also see a dx, du, dt or some other such d telling me where the scope of the integration ends, and what variable is getting integrated. In addition the integrand has to be correctly bracketed. By this I mean that, for example, writing 2x + x 2 dx ( or worse 2x + x 2 ) is wrong; it should be (2x + x 2 )dx. I know all this makes me sound like a fuddy duddy, but mathematics is a delicate instrument. If you don t treat it with respect, it becomes useless. Writing with some care and precision is part of treating the instrument with respect. End of the Note.. Chapter 6, Review, #8. Find the volume of the solid obtained by rotating the region bounded by the curves y = e 2x, y = + x, x =, about the x-axis. Solution. By slices perpendicular to the axis of rotation. A cross section at x is an annulus (washer) of inner radius e 2x, outer radius + x; its area is thus A(x) = π(( + x) 2 e 4x ). The volume we are looking for is: ( 25 V = π (( + x) 2 e 4x ) dx = π 2 + ) 4 e 4. 2. Chapter 6, Review, #6. Let R be the region in the first quadrant bounded by the curves y = x 3 and y = 2x x 2. Calculate the following quantities: (a) The area of R. (b) The volume obtained by rotating R about the x-axis. (c) The volume obtained by rotating R about the y-axis. Solution
(a) The curves intersect at x = and at x =. A picture shows that the second curve is above the first one. A = (2x x 2 x 3 ) dx = 5 2. (b) We use the method of slices. The area of a cross section perpendicular to the x-axis at x is Thus A(x) = π ( (2x x 2 ) 2 (x 3 ) 2) = π(4x 2 4x 3 + x 4 x 6 ). V = π (4x 2 4x 3 + x 4 x 6 ) dx = 4 5 π. (c) We use the method of shells. The shell at distance x from the the axis of rotation (the y-axis) has height h(x) = 2x x 2 x 3 ; its volume is dv = 2πxh(x) dx = 2πx(2x x 2 x 3 ) dx. The volume is V = 2π x(2x x 2 x 3 ) dx = 3 3 π. 3. Chapter 6, Review, #24. The base of a solid is the region bounded by the parabolas y = x 2 and y = 2 x 2. Find the volume of the solid if the cross sections perpendicular to the x-axis are squares with one side lying along the base. Solution. The parabolas intersect at x = ±. The side of the square cross sections at x is (2 x 2 ) x 2 = 2( x 2 ). It follows that V = 4 ( x 2 ) 2 dx = 4 ( 2x 2 + x 4 ) dx = 64 5. 4. Chapter 6, Review, #28. A 6 lb elevator is suspended by a 2 ft cable that weighs lb/ft. How much work is required to raise the elevator a distance of 3 ft? Solution. Let us fix the x-axis along the cable of the elevator, with the origin at the point of suspension, the elevator at x = 2 originally. In lifting it 3 feet, every part of the cable which is 3 feet or more away from the point of suspension, plus the elevator, gets moved 3 feet. We thus have 2 3 = 7 feet of cable, weighing 7 = 7 lb plus the 6 lb elevator, a total of 33 lb, being moved 3 feet. The corresponding work is 3 33 = 99 ft-lb. Now we have to see what happens with the first 3 feet of cable. An infinitesimal portion at x feet from the suspension point of length dx has weight dx and gets moved (of course) x-feet; the corresponding work is dw = x dx. Adding all up gives the work: W = 99, + 3 x dx = 3, 5 (ft-lb). 2
5. Chapter 6, Review, #32. Let R be the region bounded by y = x 2, y = and x = b, where b >. Let R 2 be the region bounded by y = x 2, x = and y = b 2. (a) Is there a value of b such that R and R 2 have the same area? (b) Is there a value of b such that R sweeps out the same volume when rotated about the x-axis? and about the y-axis? (c) Is there a value of b such that R and R 2 sweep out the same volume when rotated about the x-axis? (d) Is there a value of b such that R and R 2 sweep out the same volume when rotated about the y-axis? Solution. I may try to include a picture, but one good way of minimizing computations for this problem is to notice that R plus R 2 forms a rectangle of base b (the interval [, b] on the x-axis), height b 2. The parabola y = x 2 splits this rectangle into a lower region R, an upper region R 2. The area of the rectangle is A R = b 3. Rotating the rectangle about the x-axis produces a cylinder of base of radius b 2, height b, thus volume V x = π(b 2 ) 2 b = πb 5. Rotating the rectangle about the y-axis produces a cylinder of base of radius b, height b 2, thus volume V y = π(b) 2 b 2 = πb 4. Now let s answer the questions. (a) The area of R is thus the area of R 2 is A = b x 2 dx = b3 3, We are asking whether we can have A 2 = A R A = b 3 b3 3 = 2b3 3. b 3 3 = 2b3 3, which is impossible for b >. The answer is NO. (b) Rotating R about the x-axis produces a solid of volume V x = π b 3 x 4 dx = b5 5 π.
About the y-axis, V y = 2π b x 3 dx = b4 2 π. Equating (and using b > ) gives b = 5/2. The answer is YES, for b = 5 2. (c) We already saw that the volume swept out by R is V x computed above. The volume swept out by R 2 will then be the volume of the cylinder swept out by the rectangle mentioned above minus the volume swept out by R ; that is V 2x = V x V x = b 5 π b5 5 π = 4b5 5 π. This can never equal V x = (b 5 π)/5. The answer is NO. (d) Proceeding similarly to the last part, we notice that the question can be rephrased to; Is there a b such that V y = V y V y ; that is, b 4 2 π = πb4 b4 2 π. The perhaps surprising answer is that the two volumes are equal for all values of b. So the answer is, of course, YES. 6. Chapter 7., #48. Find the area of the region bounded by the curves y = 5 ln x and y = x ln x. Solution. The curves intersect at x = and x = 5. The only region bounded by the curves (and by nothing else) is the region {(x, y) : x 5, x ln x y 5 ln x}. The area is A = 5 The integral(s) is easily done by parts: 5 5 ln x dx = x ln x (5 ln x x ln x) dx. 5 5 x ln x dx = 5 2 x2 ln x 2 x dx = 5 ln 5 4, x 5 Putting it all together, the area works out to x 2 x dx = 25 2 ln 5 6. A = 5(5 ln 5 4) ( 25 2 ln 5 6) = 25 2 ln 5 4. Extra Credit Problems. 4
I think these are interesting problems, and it makes me sad that nobody did much with them. Problem of Chapter 6, Plus, the clepsydra problem, is one I find particularly interesting, illustrating how mathematics, albeit quite elementary mathematics, can allow you to improve designs. Maybe we ll have something similar in the exam? 7Chapter 6, Plus, #4. I refer to the textbook for the formulation of the problem and the pictures. (a) Set up the x-axis so it coincides with the diameter of the base that starts at the point where the water depth is and goes to the opposite side. Place the origin at the center of the base. A bit of reflection shows that if we slice the cylindrical glass by planes perpendicular to this x-axis, the cross sections will be rectangles. It is also not too hard to see that if we are at a point x, where x ranges from r to r, then the base of the rectangle (which lies on a line perpendicular to the x-axis, on the bottom of the glass) is 2 r 2 x 2 while the height is L(x + r)/(2r). The integral giving the volume thus works out to r (x + r)l V = r2 x2 dx = L r (x + r) r r 2r r 2 x 2 dx. r (b) To get trapezoids, we slice by planes perpendicular to the previous ones. With the x-axis as in (a), we slice by planes perpendicular to the y-axis (set up also the y-axis; it goes through the bottom of the glass and is perpendicular to the x-axis). If we take a plane at y, perpendicular to the y axis, it will intersect the circumference of the base at two points. One of them is ( r 2 y 2, y), the other one ( r 2 y 2, y). At the first point, where x = r 2 y 2, the water goes up, along the plane in question, for a length of L(x + r) 2r = L(r r 2 y 2 ). 2r This is the length of the shorter side of the trapezoid. At the second point, where x = r 2 y 2, the water goes up, again along the plane in question, for a length of L(x + r) = L(r + r 2 y 2 ). 2r 2r This is the length of the shorter side of the trapezoid. The base of the trapezoid stretches between the two x values, hence equals 2 r 2 y 2 giving A(y) = 2 ( r 2 y 2 L(r r 2 y 2 ) + L(r + ) r 2 y 2 ) = L r 2 2r 2r 2 y 2. The volume, by this method, is thus given by V = L r r r2 y 2 dy. 5
(c) The easiest integral to evaluate is the one in part (b); we get V = L r r r2 y 2 dy = πr2 L 2. Actually, the integral in part (a) is also quite easy; V = L r r r(x + r) r 2 x 2 dx = L r r r r r 2 x 2 dx and after cancelling an r, we get the same value as before. (c) It takes a bit more of reflection to realize that if we straighten the glass, the water will be at depth L/2, giving a volume of V = πr 2 L/2. (d) Some other time. 8. Chapter 6, Plus, # 6. I refer to the textbook for the statement, the pictures, and the interpretation of the variables. The exercise is actually much easier than might seem at first glance. (a) Let us suppose someone asks us to figure out the percentage of the volume of an object above the surface of the water. That is quite easy, I think. If we write V to denote the volume of the object, V a to denote the volume above the water, then the percentage is p = V a V. But how to find out anything about V, V a here? Well,we know, from what we learned in the course so far that V = L h h A(y) dy, V a = L h A(y) dy. We see at once, maybe, that V = W/(ρ g); what about V a? If we realize that L h V a = A(y) dy = V A(y) dy, h we might also realize that V a = V (F/(ρ f g)) = W/(ρ g) (F/(ρ f g)), hence p = = ρ f W ρ F ρ f W. W ρ F g ρ f g W ρ g And we might be stuck here, except that we haven t used any physics yet! The fact that the object is floating means that the buoyant force F equals the weight W ; using W = F we get as we were supposed to. p = ρ f ρ ρ f, 6
(b) This is just a question of plugging in the values. The percentage is 3 97.97. 3 So slightly less than % of the iceberg is above water. (c) This is actually a very nice question which only needs the principle of Archimedes to be answered. Here is a straight answer, ignoring what we did here so far. The ice is floating in the water which means that the portion of the ice in the water has displaced a volume of water weighing exactly what the ice cube weighs. But once the ice cube has melted, it is the same as the water in which it floats, so it will replace exactly the water it has displaced. The water does not overflow when the ice cube melts, nor does the level fall. A solution using part (a) could go as follows: Once the ice melts, one has ρ f = ρ, giving a percentage of water above water of %. I prefer the previous answer. (d) Since the sphere has negligible weight, it is % above the water (We assume that its weight is not so negligible that it also floats up into the wild blue yonder, but instead assume that it is just touching the water). We can think of this as follows: We are going to push the sphere down, at each stage we have to counteract a force which is equal to the weight of the submerged portion of the sphere. Suppose the sphere has been submerged h meters (measured in a direction perpendicular to the surface of the water), what is the submerged volume? If we place a y axis with the origin at the center of the sphere, we are asking for the volume of the portion of the sphere between y =.4 and y =.4 + h, which works out to.4+h V (h) = π (.4) 2 y 2 ) dy = π(.4h 2 h3 3 )..4 (If we slice the sphere at y, a bit of geometry the theorem of Pythagoras shows us that we get a circle of radius (.4) 2 y 2, surface area π ( (.4) 2 y 2). The infinitesimal slice at y is an infinitesimal cylinder of height dy, thus volume dv = π ( (.4) 2 y 2) dy.) That means that when the sphere has been submerged h meters, the force acting on it (the buoyancy force) is ) F = g mass = g density V (h) = πg (.4h 2 h3 3 ). The work is thus W =.8 F (h) dh = πg.8 which works out to approximately 5.88 Joules. ) ( (.4h 2 h3.248 3 ) = πg 3.496 ) 2 7
9. Chapter 6, Plus, #. I refer to the textbook for the formulation of the problem and the pictures. (a) The method of slices gives V (h) = π h f(y) 2 dy. (b) By the chain rule and the fundamental theorem of Calculus, ( dv dt = dv dh dh dt = d ) h π f(y) 2 dh dh dy = πf(h)2 dh dt dt. (c) If dh/dt = C is constant, then from Torricelli s law and part (b), Solving, we get πcf(h) 2 = ka h. f(h) = ( ) /2 ka h /4, Cπ or, since C will be absorbing all other constants, and the square root of a constant is just a constant, f(h) = Ch /4. The advantage of having dh/dt constant is that the height of water decreases at a rate proportional to the rate of change in time. For example, you can arrange (by choosing C appropriately) that a decrease of one inch in the water level corresponds to one hour; then a decrease of one inch will be an hour whether the vessel is full or almost empty.. Chapter 7. #64. Some parts of this exercise are very easy, some are not. For example, since sin x in the interval [, π/2], it follows that sin k+ x = sin k x sin x sin k x in that interval; specializing to k = 2n and k = 2n + one gets sin 2n+2 x sin 2n+ x sin 2n x; integrating from to π/2 gives I 2n+2 I 2n+ I 2n. That takes care of part (a). Now Exercise 4 gives you at once I 2n+2 I 2n = which takes care of part (b). Now 3 5 (2n+) π 2 4 6 (2n+2) 2 3 5 (2n ) π 2 4 6 2n 2 = 2n + 2n + 2 (2n + ) (2n + 2) = I 2n+2 I 2n I 2n+ I 2n, 8
the inequalities being immediate consequences of part (a). Thus I 2n+ lim = n I 2n by the squeeze Theorem. And part (c) is done. (d) This is perhaps the ONLY difficult part in this exercise. Doing it would have been good practice in problem solving. How does one do something like this? The first thing to do is to get familiar with the problem, try to get some idea of how things go. Bring math in. Since we are repeating a certain operation again and again, we might try to sort of label the diverse steps by the positive integers more or less as follows. Step. Draw the square. The width and height are, Step 2. Add a rectangle to the right. Step 3. Add a rectangle on top. Step 4. Add a rectangle to the right. Step 5. Add a rectangle on top....... How do the widths and heights of the figures change? Let us denote the width of the figure of the n-th step by w n, the height by h n. Now, it should be clear that everytime we add a rectangle to the right its width should be the reciprocal (one over) the height of the figure to which the rectangle is added; when adding a rectangle at the top, its height should be the reciprocal of the figure s width. That s needed so the area of the added rectangle is. We start with h =, w = and get the following list of new widths, heights: w =, h =, w 2 = w + h = 2, h 2 = h =, w 3 = w 2 = 2, h 3 = h 2 + w 2 = 3 2, w 4 = w 3 + h 3 = 2 + 2 3 = 8 3, h 4 = h 3 = 3 2, w 5 = w 4 = 8 3, h 5 = h 4 + w 4 = 3 2 + 3 8 = 5 8, w 6 = w 5 + h 5 = 8 3 + 8 5 = 48 5, h 6 = h 5 = 5 8, w 7 = w 6 = 48 5, h 7 = h 6 + w 6 = 5 8 + 5 48 = 5 48, 9
How long should one go on? Until one sees a pattern, is the answer. It helps to keep the previous parts of the exercise in mind, since you may suddenly realize that one has w 2n+ = w 2n = h 2n+2 = h 2n+ = 2 4 (2n) 3 (2n ), 3 (2n + ) 2 4 (2n). (Products like these showing up in parts (a), (b), (c), (d)). If one is to be % precise and careful, one should make sure that these formulas are right; that is done by mathematical induction and I ll skip it. So I assume the formulas conjectured for the width and height have been established. Comparing with Exercises 39, 4, we see that it follows that and w 2n+ = w 2n = (2n + ) π/2 h 2n+2 = h 2n+ = (2n + ) 2 π From this it is immediate that sin 2n+ x dx = (2n + )I 2n+, π/2 w 2n = (2n + )I 2n+ h 2n (2n ) 2 π I = π 2n 2 2 sin 2n x dx = (2n + ) 2 π I 2n, (2n + ) I 2n+, (2n ) I 2n 2 w 2n+ = (2n + )I 2n+ h 2n+ (2n + ) 2 π I = π I 2n+. 2n 2 I 2n w n lim = π n h n 2.