AlphaMaster From: US Posts: 26 Votes: 52 |
Find the volume of the solid obtained by rotating the region bounded by y = x2 and y = √x adout the line x=2. |

Anonymous From: - Posts: - Votes: - |
Hi, this is a same question with just different numericals . I am posting both question and answer. Please rate me Lifesaver !!!! Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y=(1/x), y=0, x=1, x=3; about y=-1 ANS:Shell method: V = 2p ? r h dr {a,b} .......... limits in {} radius is (1+y) to rotate around y=-1 = 2p ?(1+y)(2) dx {0,1/3} + 2p ?(1+y)(1/y - 1) dx {1/3,1} = 4p [y + y²/2] {0,1/3} + 2p [ ln(y) - y²/2 ] {1/3,1} = 14p/9 + 2p [ ln(3) - 4/9 ] = 2p [ ln(3) + 1/3 ] ˜ 8.997 units³ Disc method: V = p ? f(x)² dx {a,b} rotate around y=-1 extends the inner and outer radii by 1 = p ? (1/x + 1)² - (1)² dx {1,3} .............. note: p [(Ro)² - (Ri)²] = p ? 1/x² + 2/x dx {1,3} = p [ -1/x + 2 ln(x) ] {1,3} = p [ -1/3 + 2 ln(3) + 1 ] = 2p [ ln(3) + 1/3 ] Answer: ˜ 8.997 units³ |

Anonymous From: - Posts: - Votes: - |
Similar one :) curve: y-sqrt of x lines: y=2 x=0 if we will revolve this about the y-axis what would be its resulting solid of revolution.how would this look like,,? y = vx, so x = y² You will integrate with respect to y from 0 to 2. Your discs will have area A = pi r², where r = x = y², so A = pi y^4 Integrate from 0 to 2 ?pi y^4 dy = (2/5) pi y^5 evaluated from 0 to 2 64pi/5 |

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