dominet Posted October 21, 2011 Posted October 21, 2011 (edited) Hello, I'm trying to find the Volume of this solid by the Shell Method but can't figure out what the height of the shell is for this: rotation about x-axis of region bounded by y=x^2 and y=6x+7 the formula should be: Volume= 2pi int(radius*height) dy from 0 to 49. I know radius is y. But what is the height? Edited October 21, 2011 by dominet
Xittenn Posted October 21, 2011 Posted October 21, 2011 (edited) -- sorry, I misunderstood -- Edited October 21, 2011 by Wuz Here
dominet Posted October 21, 2011 Author Posted October 21, 2011 (edited) Annnd I just answered my own post again: divide the volume into 2 volumes (first from y=0 till y=1, second from y=1 till y=49) then add them. In other words: http://www.wolframal...+dy+from+0+to+1 ++++Plus++++ http://www.wolframal...dy+from+1+to+49 Edited October 21, 2011 by dominet
Xittenn Posted October 21, 2011 Posted October 21, 2011 (edited) Annnd I just answered my own post again: divide the volume into 2 volumes (first from y=0 till y=1, second from y=1 till y=49) then add them. In other words: http://www.wolframal...+dy+from+0+to+1 ++++Plus++++ http://www.wolframal...dy+from+1+to+49 gratz Briggs & Cochran Calculus: Early Transendentals gives a good description of the shell method as well! [math] V = \int^b_a 2 \pi x ( f(x) - g(x) ) dx [/math] Edited October 21, 2011 by Wuz Here
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