Revolution Disc
Revolution Disc
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1 KONAMI DANCE DANCE REVOLUTION GAME DISC CD ONLY, FOR NINTENDO WII $12.71 |
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Rotaz Revolution Disc Hubs,381g,Golden,32H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,381g,Red,28H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,381g,Red,32H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,Black,381g,28H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,Black,381g,32H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,White,381g,32H,Titanium QR $139.95 |
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Rotaz Revolution Disc Hubs,Black,381g,28H,Ti Skewer $154.95 |
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Rotaz Revolution Disc Hubs,Black,381g,32H,Ti Skewer $154.95 |
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Rotaz Revolution Disc Hubs,381g,Red,32H,Ti Skewer $154.95 |
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Rotaz Revolution Disc Hubs,381g,Red,28H,Ti Skewer $154.95 |
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STAN’S ZTR RACE WHEELS / AMERICAN CLASSIC DISC HUBS / DT REVOLUTION – 1190g! $500.00 |
Could you help me solve a problem of solids of revolution?
I've been dealing with a problem for a few ... well ... hours now. Since yesterday afternoon. Even now I could not get it. I wonder if someone could explain how this problem be solved? Find the volume of the region bounded by y = x 2, y =- x-2 x = 0 turn around (a) and =- 2, (b) x = -2 using the best method available (washers, discs, or cylindrical shells). Thanks in advance, Sam.
a) It seems obvious to me that the best method to solve this problem is with the method of the washing machine. The problem comes with the correct equations to describe the change of the radii of the circles forming the shape of the washer. If you draw a graph of this monster, a side view reveals that what is really doing is rotating a pyramid shape about an axis, in this case y = -2, which will in a circular pyramid, so to speak. Let's do some quick comments on the circles produce washer as we will integrate areas for find the volume. First, starting from x = -2 and x = 0, the radius at any given point of one of the largest circles is equal to any arbitrary value and the equation y = x + 2 minus the center of the circle, y = -2. Therefore y = R - (-2) = y + 2 = (x + 2) + 2 = x + 4. The radius, r, of each small circle is given by y - (-2) y = '+ 2 = (-x - 2) + 2 =- x, where y is an arbitrary value and the line equation y = - x - 2. Then, r =- x. Note however that since all values of x here, except 0, is negative, then r = - (-x) = + x. Whatever the method of the washing machine does is simply subtracting the area a circle from the area in a larger circle. In this case, the larger circle is the one whose radius is R, and the smaller circle has radius r. So that the resulting difference in areas is given by the following equation: A = Π R ² - π r ². Excluding π, we get: A = π (R ² - R ²). Now we just have to connect our values of R and R, and squares: A = π [(x + 4) ² - (x)] ². A = π (x ² + 8x + 16 - x ²) = π A (8x + 16). Now we can simply integrate the above equation along the axis x term by term between the limits x = -2 x = 0 for find the volume of the solid. V = ∙ π [Integral x = -2 to 0 (8x + 16)] dx = V π ∙ 4x [+ 16x ² | -2 to 0] V = π ∙ [4 (0) ² + 16 (0) - [4 (-2) ² + 16 (-2) V] = π ∙ [0 + 0 - (1916-1932)] V = π ∙ [0 - (-16)] = V ∙ π (16) V = π 16 cubic units so that the solid has a volume of 16 π cubic units b) The two solids represented by this rotation are really only two cones consistent coincide with vertices at x = -2. The upper cone is inverted in the inferior one. The easiest way to integrate this is to use the method of disc. Let's find the area of a representative disk oriented horizontally, and then integrate along the axis. For any arbitrary disk formed by this cone, the radio and in any given point along the line y = x + 2 from y = 0 and y = 2 is given by r = y = x + 2. Therefore, when y = 0, x = -2, r = -2 + 2 = 0, and when y = 2, x = 0 and r = 0 + 2 = 2. So the area of each of these discs can be given by this formula: A = π r ². The volume of the solid areas composed of all disks is the following: V = Integral ² π and d and y = 0 and y = 2. V = π (and ³ / 3) | 0-2 V = π [(2) ³ / 3 - (0) ³] V = π (8.3) V = (8.3) π from the lower cone is consistent with the upper cone, double the volume just above to find the total volume. V (Total) = 2 [(8.3)] π V (total) = (16 / 3) π we find the volume of two cones also using the shell method, but with more difficulty. Remember shell method is like building layers and layers of thin sheets of paper wrapped around each other inside out. If you are wrapped in a circular shape Then the circumference of each leaf so involved is given by C = 2π r, where r is given by the equation x 2 + Therefore, when x = -2 r = 0, and when x = 0, = r 2. Therefore, C = 2π (2 + x). The height, h, of each leaf decreases as one moves from x = -2 and x = 0, and is given by h = 2 - y, where y = x + 2. Then h at any point is 2 - (x + 2) = 2-2 - x =- x. The total area of each leaf is equal to: A, C h ∙ A = = [2π (2 + x)] (-x) A = 2π (-2x - x ²) To find the volume of the upper cone, we integrate the area formerly used along the x-axis to get this: V = 2π [Integral (-2x - x ²) x = -2 and x = 0] V = 2π (-x ² - x ³ / 3) | -2 A 0 V = 2π (-0 ² - 0 ³ / 3) - [- (-2) ² - (-2) ³ / 3 V] = 2π [0 - (-4 + 8 / 3) V] = 2π [- (-12 / 3 + 8 / 3)] V = 2π [- (-12 + 
/ 3] V = [2π - (-4 / 3)] V = 2π (3.4) V = (3.8) π To calculate the volume of two cones, multiply the figure by over a 2. So the total volume using this method is the same as the disk method: V (total) = 2 [(8.3) π] V (total) = (16 / 3) π.