How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=6sin(5x^2), between 0 and sqrt(pi/5) revolved about Ox?

1 Answer
Dec 10, 2016

2.4pi (7.54 2 dp)

Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness deltax between the x-axis and the curve at some particular x-coordinate it would have an area:

delta A ~~"width" xx "height" = ydeltax = f(x)deltax
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If we then rotated this infinitesimally thin vertical line about Oy then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume delta V would be given by:

delta V~~ 2pi xx "radius" xx "thickness" = 2pixdeltaA=2pixf(x)deltax

If we add up all these infinitesimally thin cylinders then we would get the precise total volume V given by:

V=int_(x=a)^(x=b)2pixf(x) dx

So for this problem we have:

\ \ \ \ \ V = int_0^sqrt(pi/5) 2pix(6sin(5x^2)) dx
:. V = 12piint_0^sqrt(pi/5) xsin(5x^2) dx
:. V = 12pi [-cos(5x^2)/10]_0^sqrt(pi/5)
:. V = (-12pi)/10 [cos(5x^2)]_0^sqrt(pi/5)
:. V = (-12pi)/10 (cospi-cos0)
:. V = (-12pi)/10 (-1-1)
:. V = (24pi)/10
:. V = 2.4pi (~~7.5398...)