2 - Solutions - MATH 1080 Calculus II Homework#2 Name_SOLUTIONS Group Table Due Monday January 27th(beginning of class Topics Covered Section 6.3 Volume

2 - Solutions - MATH 1080 Calculus II Homework#2...

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Unformatted text preview: MATH 1080, Calculus II Homework #2 Name: ________SOLUTIONS_________ Group: __________ Table: __________ Due: Monday, January 27th (beginning of class) Topics Covered: • Section 6.3: Volume by Slicing & Disk/Washer Method • Section 6.4: Volume by Shell Method Directions: Answer the following questions. Do all your work on your own paper. When you are finished, staple all your work to this sheet. To receive full credit you must show all of your work and provide 1) Proper and complete notation!! 2) Legible, organized, and logical (relevant) justification for your answer!! 3) Each question should be concluded with a complete sentence, either mathematical or in words as appropriate!! Conceptual Questions: 1. True/False: Determine whether the following statements are true or false. If true, give an explanation; if false, give a counterexample. a. When using the shell method, the line segment used to represent a typical shell is parallel to the axis of revolution. Solution: True, if the line segment were not parallel, then rotating it around the axis of revolution would not result in a shell. b. If a region is revolved about the y-axis, then the disk method must be used. Solution: False, in principle either disk/washer or shell methods can be used for revolution around the y-axis. c. If a region is revolved about the x-axis, then in principle it is possible to use the disk/washer method and integrate with respect to x or the shell method and integrate with respect to y. Solution: True, either method can be used for revolution around the x-axis, and revolution around the x-axis would require x for the disk/washer method and y for the shell method. 2. Explanation: a. Suppose a cut is made through a solid object perpendicular to the x-axis at a particular value of x. Explain the meaning of A(x). Solution: A(x) is the area of a typical cross section that is cut through the solid at the particular value of x. b. Explain why the disk method is a special case of the general slicing method. Solution: For the disk method, the cross sections are disks or circles, and the area of the cross section A(x) is the area of a circle, ߨሾ‫ݎ‬ሺ‫ݔ‬ሻሿଶ . c. A solid has a circular base and cross sections perpendicular to the base are squares. What method should be used to find the volume of the solid that is generated? Explain why the method you have chosen must be used and why another method cannot be used. Solution: The general slicing method must be used because the cross-sections are square. Because of the square crosssections, the solid would not be able to be formed by rotation about an axis thereby ruling out the disk/washer and shell methods. Problem Solving Questions: 3. Use the method of slicing to compute the volume, in cubic feet, of the Great Pyramid of Egypt, whose base is a square 755 feet by 755 feet and whose height is 410 feet. Note: your final answer will be large; therefore, you can use a calculator to find your final answer, and you can round your final answer to two significant digits. However, you must show all of your work up until the last step or arithmetic. Solution: We slice the pyramid horizontally. Each slice is a square with area ‫ ݏ‬ଶ . To find a formula for ‫ ,ݏ‬a side of the square, we will draw a triangle in the xy-plane to represent the silhouette of the pyramid with a horizontal slice in the middle (see below). We will place the x-axis on the bottom of the triangle and the y-axis in the middle of the triangle. The left side of the triangle is a line with ଼ଶ଴ ଵହଵ ଻ହହ equation ‫ ݕ‬ൌ ‫ ݔ‬൅ 410 ⇒ ‫ ݔ‬ൌ ‫ ݕ‬െ , and the right side is a ଻ହହ line with equation ‫ ݕ‬ൌ 410 െ ଼ଶ଴ ଻ହହ ଵ଺ସ ‫ݔ⇒ݔ‬ൌ ଶ ଻ହହ ଶ െ ଵହଵ ଵ଺ସ ‫.ݕ‬ ‫ݕ‬ ‫ݔ‬ Then ‫ ݏ‬ൌ ݀‫ ݑ‬ൌ െ ଻ହହ ଶ ଵହଵ െ ଵହଵ ଵ଺ସ ‫ݕ‬െቀ ‫ݕ‬െ ଵହଵ ݀‫ ⇒ ݕ‬െ ଼ଶ ଵ଺ସ ଻ହହ ଶ ቁ ൌ 755 െ ଵହଵ ଼ଶ ‫ .ݕ‬Therefore, ܸ ൌ ‫׬‬ ଴ ସଵ଴ ቀ755 െ ଵହଵ ଼ଶ ‫ݕ‬ቁ ݀‫ ݑ ݃݊݅ݐݐ݈݁ ݓ݋ܰ .ݕ‬ൌ 755 െ ଶ ݀‫ ݑ‬ൌ ݀‫ ݕ ݄݊݁ݓ ݋ݏ݈ܣ .ݕ‬ൌ 0, ‫ ݑ‬ൌ 755; ‫ ݕ ݄݊݁ݓ‬ൌ 410, ‫ ݑ‬ൌ 0. ݄ܶ݁‫,݁ݎ݋݂݁ݎ‬ ଵହଵ ଼ଶ ‫⇒ ,ݕ‬ ଴ 151 ଶ 82 ଶ 82 ଻ହହ ଶ 82 ଷ ଻ହହ 82 ሺ755ሻଷ ൎ 78 ݈݈݉݅݅‫ ݐ݂݁݁ ݊݋‬ଷ . ܸ ൌ න ൬755 െ ‫ݕ‬൰ ݀‫ ݕ‬ൌ න െ ‫ ݑ݀ ݑ‬ൌ න ‫ ݑ݀ ݑ‬ൌ ൤ ‫ ݑ‬൨ ൌ 82 151 151 ଴ 453 453 ଴ ଴ ଻ହହ Consequently, the volume of the great pyramid is approximately ܸ ൎ 78 ݈݈݉݅݅‫ ݐ݂݁݁ ݊݋‬ଶ . ଼ଶ ସଵ଴ ଵହଵ 4. A solid has a base in the ‫-ݕݔ‬plane that is bounded by the curves ‫ ݕ‬ൌ ‫ ݔ‬ଶ and the line ‫ ݕ‬ൌ 1. Cross sections through the solid perpendicular to the ‫-ݕ‬axis are triangles that are twice as tall as their base. Find the volume of the solid. Solution: The cross sections are triangles therefore, ‫ ܣ‬ൌ ܾ݄ ൌ ܾሺ2ܾሻ ൌ ܾ ଶ . ଶ ଶ Now the base of the triangle is perpendicular to the ‫ ݕ‬axis running from right side of the parabola to the left side (see figure to right) – therefore we need the parabola to be in terms of ‫ ݕ :ݕ‬ൌ ‫ ݔ‬ଶ ⇒ ‫ ݔ‬ൌ േඥ‫ .ݕ‬Then ܾሺ‫ݕ‬ሻ ൌ ඥ‫ ݕ‬െ ଵ ଵ ൫െඥ‫ݕ‬൯ ൌ 2ඥ‫ .ݕ‬Therefore, ‫ ܣ‬ൌ ൫2ඥ‫ݕ‬൯ ൌ 4‫ .ݕ‬Also, ‫ ݕ‬ranges from 0 to 1 so ଶ ܸ ൌ න 4‫ ݕ݀ݕ‬ൌ ሾ2‫ ݕ‬ଶ ሿଵ ൌ 2 ‫ ݏݐ݅݊ݑ‬ଶ . ଴ ଵ Consequently, the volume of the solid is ܸ ൌ 2 ‫ ݏݐ݅݊ݑ‬ଶ . ଴ ଵ ௫ 5. Washers vs. Shells: Let ܴ be the region bounded by ‫ ݕ‬ൌ and ‫ ݕ‬ൌ 1 െ . Let ܵ be the solid generated ௫ାଵ ଷ when ܴ is rotated about the ‫-ݔ‬axis. Find the volume of ܵ by both the disk/washer and shell methods. Check that your results agree. State which method is easiest to use and why. Solution: First, we must find the points of intersection: ‫ݔ‬ 1 3െ‫ݔ‬ 1 ൌ 1െ ⇒ ൌ ⇒ 3 ൌ ሺ3 െ ‫ݔ‬ሻሺ‫ ݔ‬൅ 1ሻ ⇒ 0 ൌ 2‫ ݔ‬െ ‫ ݔ‬ଶ ⇒ 0 ൌ ሺ2 െ ‫ݔ‬ሻ‫ ݔ ⇒ ݔ‬ൌ 0,2. ‫ݔ‬൅1 3 ‫ݔ‬൅1 3 ଵ Then the two points of intersection are ሺ0,1ሻ and ቀ2, ቁ. ଷ For the disk/washer method, rotation about the ‫-ݔ‬axis implies integration with respect to ‫ .ݔ‬Graphing the region described shows that there is a gap between the region and the axis of revolution; therefore, we use the washer method with ଵ ௫ ‫ݔ‬ ‫ݎ‬ሺ‫ݔ‬ሻ ൌ and ܴሺ‫ݔ‬ሻ ൌ 1 െ . Then ‫ ݕ‬ൌ 1െ ‫ ݔ‬ଶ 1 ଶ ܸ ൌ න ߨ ቈቀ1 െ ቁ െ ൬ ൰ ቉ ݀‫ݔ‬ 3 ‫ݔ‬൅1 ଴ ଶ ଶ 2 1 1 ൌ ߨ න ൬1 െ ‫ ݔ‬൅ ‫ ݔ‬ଶ ൰ ݀‫ ݔ‬െ ߨ න ݀‫ݔ‬ ሺ‫ ݔ‬൅ 1ሻଶ 3 9 ଴ ଴ ᇣᇧᇤᇧᇥ ௫ାଵ ଶ ଷ 3 ‫ݕ‬ൌ 1 ‫ݔ‬൅1 ଷ 1 ଷ ଶ 4 8 1 ଷ ‫ ݔ‬൨ െ ߨ න ‫ିݑ‬ଶ ݀‫ ݑ‬ൌ ߨ ൤2 െ ൅ ൨ െ ߨሾെ‫ିݑ‬ଵ ሿଵ ൌ ߨ ൤‫ ݔ‬െ ‫ ݔ‬ଶ ൅ 3 27 3 27 ଴ ଵ 26ߨ 1 8ߨ ൌ െ ߨ ൬െ ൅ 1൰ ൌ ‫ ݐ݅݊ݑ‬ଶ . 27 3 27 ௟௘௧ ௨ୀ௫ାଵ For the shell method, rotation about the ‫-ݕ‬axis implies integration with respect to ‫ .ݕ‬Therefore, solving both functions for ‫ݕ‬ gives: 1 1 ‫ݔ‬ ⇒ ‫ ݔ‬ൌ െ 1 and ‫ ݕ‬ൌ 1 െ ⇒ 3 െ 3‫ ݕ‬ൌ ‫.ݔ‬ ‫ݕ‬ൌ ‫ݔ‬൅1 ‫ݕ‬ 3 Now as the axis of revolution is the ‫-ݔ‬axis, the shell radius is ‫ ;ݔ‬also from the graph of the region, it is clear that the shell height ଵ is 3 െ 3‫ ݕ‬െ ቀ െ 1ቁ. Consequently, ௬ ଵ ଵ 1 8ߨ ܸ ൌ න 2ߨ‫ ݕ‬൭3 െ 3‫ ݕ‬െ ൬ ൅ 1൰൱ ݀‫ ݕ‬ൌ 2ߨ න ሺ4‫ ݕ‬െ 3‫ ݕ‬ଶ െ 1ሻ݀‫ ݕ‬ൌ 2ߨሾ2‫ ݕ‬ଶ െ ‫ ݕ‬ଷ െ ‫ݕ‬ሿଵ ൌ ‫ ݏݐ݅݊ݑ‬ . ଵ/ଷ ଵ ଵ ‫ݕ‬ 27 ଷ Consequently, ܸ ൌ ଷ ଼గ ଶ଻ ‫ ݏݐ݅݊ݑ‬ଶ from both methods. Since both methods resulted in an integral that could be evaluated using our current integration techniques, an argument could be made for either method being the easier option. The washer method may be considered the easier option since it didn’t require solving for the other variable. However, the integration in terms of y was much simpler, so an argument could also be made for the shell method being the easier option in this case. 6. Let ܴ be the region bounded by the following curves. Let ܵ be the solid generated when ܴ is revolved about the given axis. Find the volume of ܵ by the method of your choice; in your solution, state the method that you use. ଵ a. ‫ ݕ‬ൌ cosሺ‫ ݔ‬ଶ ሻ , ‫ ݔ‬ൌ 0, ‫ ݔ‬ൌ √ߨ, ‫ ݕ‬ൌ 0, about the ‫-ݕ‬axis. ଶ Solution: Notice, rotation around the ‫-ݕ‬axis requires disk/washer method to be in terms of ‫ .ݕ‬However, if we solve for ‫ ݕ‬in the first equation, we obtain ‫ ݔ‬ൌ ඥܿ‫ି ݏ݋‬ଵ ‫ ݕ‬which we cannot integrate. Therefore, we must integrate with respect to ‫ .ݔ‬Consequently, we must use the shell method. Then ଵ ଶ√గ ܸൌන ଴ గ ସ 1 ‫ ݔ‬ൌ √‫ݔ‬ 2 ‫ݕ‬ൌ0 2ߨ‫ݏ݋ܿݔ‬ሺ‫ ݔ‬ଶ ሻ ᇣᇧᇧᇤᇧᇧᇥ ݀‫ ݔ‬ൌ ߨ න ܿ‫ ݑ݀ ݑ ݏ݋‬ൌ ߨሾ‫ݑ ݊݅ݏ‬ሿ଴ ᇧ ᇧ గ/ସ ଴ ߨ √2 ൌ ߨ ቂ‫ ݊݅ݏ‬െ ‫0 ݊݅ݏ‬ቃ ൌ ߨ. 4 2 √ଶ Therefore, the volume of the solid is ܸ ൌ ߨ ‫ ݐ݅݊ݑ‬ଷ . ௅௘௧ ௨ୀ௫ మ ‫ݔ‬ൌ0 ଶ b. ‫ ݔ‬ൌ ‫ ݔ , 2ݕ‬ൌ െ‫ ݕ ,ݕ‬ൌ 2, ‫ ݕ‬൒ 0, about the ‫-ݔ‬axis. ‫ݕ‬ൌ2 Solution (Shell Method): Shell radius is ‫ ݕ‬and shell height is ‫ ݕ‬ଶ െ ሺെ‫ݕ‬ሻ ൌ ‫ ݕ‬ଶ ൅ ‫ .ݕ‬Then ଶ ଶ ଶ ‫ݕ‬ସ ‫ݕ‬ଷ ଶ ଷ ଶ ሻ݀‫ݕ‬ ൌ 2ߨ ቈ ൅ ቉ ܸ ൌ න 2ߨ‫ݕ‬ሺ‫ ݕ‬൅ ‫ݕ‬ሻ݀‫ ݕ‬ൌ න 2ߨሺ‫ ݕ‬൅ ‫ݕ‬ 4 3 ଴ ଴ ଴ 8 40ߨ ൌ 2ߨ ൬4 ൅ ൰ ൌ ‫ ݐ݅݊ݑ‬ଷ . 3 3 ସ଴గ Therefore, the volume of the solid is ܸ ൌ ‫ ݏݐ݅݊ݑ‬ଷ . ‫ ݔ‬ൌ ‫ݕ‬ଶ ଷ ‫ ݔ‬ൌ െ‫ݕ‬ Solution (Washer Method): First, put the equations in terms of ‫ ݔ :ݔ‬ൌ ‫ ݕ‬ଶ ⇒ ‫ ݕ‬ൌ േ‫ ,ݔ‬and ‫ ݔ‬ൌ െ‫ ݕ ⇒ ݕ‬ൌ െ‫ .ݔ‬Then ܸ ൌ ߨ න ሺ2ଶ െ ሺെ‫ݔ‬ሻଶ ሻ݀‫ ݔ‬൅ ߨ න ቀ2ଶ െ ൫√‫ݔ‬൯ ቁ ݀‫ ݔ‬ൌ ߨ න ሺ4 െ ‫ ݔ‬ଶ ሻ݀‫ ݔ‬൅ ߨ න ሺ4 െ ‫ݔ‬ሻ݀‫. ݔ‬ ଴ ସ ଶ ଴ ସ ଴ ସ 1 1 8 40ߨ ൌ ߨ ൤4‫ ݔ‬െ ‫ ݔ‬ଷ ൨ ൅ ߨ ൤4‫ ݔ‬െ ‫ ݔ‬ଶ ൨ ൌ ߨ ൤0 െ ൬െ8 ൅ ൰൨ ൅ ߨሾሺ16 െ 8ሻ െ 0ሿ ൌ ‫ ݏݐ݅݊ݑ‬ଷ . 3 2 3 3 ିଶ ଴ ସ଴గ Therefore, the volume of the solid is ܸ ൌ ‫ ݏݐ݅݊ݑ‬ଷ . ିଶ ଴ ିଶ ଴ ଷ c. ‫ ݕ‬ൌ sec ‫ ݕ , ݔ‬ൌ tan ‫ ݔ , ݔ‬ൌ 0, ‫ ݔ‬ൌ 1, about the ‫-ݔ‬axis. Solution: If we use the shell method, then we must solve for ‫ݔ‬ which gives us two inverse trig functions which we don’t know how to integrate right now. Therefore, we must integrate with respect to ‫ .ݔ‬Then ܸ ൌ න ߨሾሺ‫ݔ ܿ݁ݏ‬ሻଶ െ ሺ‫ݔ ݊ܽݐ‬ሻଶ ሿ݀‫ݔ‬ ଵ ଴ ൌ න ߨሾ‫ ܿ݁ݏ‬ଶ ‫ ݔ‬െ ሺ‫ ܿ݁ݏ‬ଶ ‫ ݔ‬െ 1ሻሿ݀‫ݔ‬ ଵ ଴ ‫ ݕ‬ൌ sec ‫ݔ‬ ଵ ൌ න ߨሺ1ሻ݀‫ ݔ‬ൌ ߨሾ‫ݔ‬ሿଵ ଴ ൌ ߨ ‫. ݏݐ݅݊ݑ‬ Therefore, the volume of the solid is ܸ ൌ ߨ ‫ ݏݐ݅݊ݑ‬ଷ . ଴ ‫ ݕ‬ൌ tan ‫ݔ‬ ଷ ‫ݔ‬ൌ0 ‫ݔ‬ൌ1 ...
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