Usually, the disc method is preferred when the graph is revolved about the x axis.<\/p>\n
Example:<\/strong> Find the volume generated, x=7 to x=10, when y=-(x-9)2<\/sup>+5 is revolved about the x axis.<\/p>\n Solution:<\/p>\n Here we have an ideal case for the disc method. The radius of each disc is the height of the graph above the x axis; each disc’s area, therefore, is \u03c0(height)2<\/sup>. Of course, the height is y, or -(x-9)2<\/sup>+5. The “thickness” of the disc is dx. Using the concept that<\/p>\n volume=(area)(thickness)<\/p>\n we construct the integral:<\/p>\n V=\u222b7<\/sub>10<\/sup>\u03c0(-(x-9)2<\/sup>+5)2<\/sup>dx<\/p>\n which becomes, from expanding the outer square,<\/p>\n V=\u222b7<\/sub>10<\/sup>\u03c0((x-9)4<\/sup>-10(x-9)2<\/sup>+25)dx<\/p>\n We can factor out \u03c0, then integrate term by term:<\/p>\n \u03c0|10<\/sup>7<\/sub> (x-9)5<\/sup>\/5 – 10(x-9)3<\/sup>\/3 + 25x<\/p>\n Plugging in for the limits of integration we get<\/p>\n \u03c0((10-9)5<\/sup>\/5-10(10-9)3<\/sup>\/3+25(10) then<\/p>\n \u03c0(1\/5-10\/3+250 -[-32\/5+80\/3+175])<\/p>\n =\u03c0(33\/5-90\/3+75)=\u03c0(33\/5+45)=\u03c0(258\/5)<\/p>\n Apparently the volume arising from the revolution described above is 258\u03c0\/5.<\/p>\n I’ll be discussing other methods of finding volumes of revolution in future posts:)<\/p>\n Source:<\/p>\n Larson, Roland and Robert Hostetler. Calculus<\/u>, third edition. Toronto: D C Heath and Company, 1989.<\/p>\n
-[(7-9)5<\/sup>\/5-10(7-9)3<\/sup>\/3+25(7)]<\/p>\n