Asked by makoy
find the volume of the solid of revolution generated when the region bounded by the curve y =x^2,the x-axis, and the lines x = 1 and x = 2 is revolved about the x-axis.
Answers
Answered by
Damon
integral from x =1 to x = 2 of 2 pi x y dx
= 2 pi integral [ x^3 dx]
= 2 pi x^4/4 from 1 to 2
= pi /2 [ 16 - 1 ] = 15/2 pi
= 2 pi integral [ x^3 dx]
= 2 pi x^4/4 from 1 to 2
= pi /2 [ 16 - 1 ] = 15/2 pi
Answered by
oobleck
The volume of a disk is πr^2 h. So, with discs of thickness dx, the volume is
∫[1,2] πy^2 dx = ∫[1,2] πx^4 dx = π/5 x^5 [1,2] = 31π/5
Or, if you want to use shells, things get a bit more complicated. The volume consists of a solid cylinder of radius1 and height 1, plus a region bounded by the curve. The volume of a cylinder is 2πrh dy. That makes the whole thing
π + ∫[1,4] 2πy(2-√y) dy = 31π/5
∫[1,2] πy^2 dx = ∫[1,2] πx^4 dx = π/5 x^5 [1,2] = 31π/5
Or, if you want to use shells, things get a bit more complicated. The volume consists of a solid cylinder of radius1 and height 1, plus a region bounded by the curve. The volume of a cylinder is 2πrh dy. That makes the whole thing
π + ∫[1,4] 2πy(2-√y) dy = 31π/5
Answered by
Damon
Whoops, sorry, I spun it around the y axis. Use oobleck reply.
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