(a) using discs of thickness dx,
v = ∫ πr^2 dx
where r = y = x(4-x)^2
v = ∫[0,4] π(x(4-x)^2) dx
(b) using shells of thickness dx, and taking advantage of symmetry,
v = 2∫[0,4] 2πrh dx
where r=x and h=y=(4-x)√x
v = 2∫[0,4] 2πx(4-x)√x dx
Consider the graph of y^2 = x(4-x)^2 (see link). Find the volumes of the solids that are generated when the loop of this graph is revolved about (a) the x-axis, (b) the y-axis, and (c) the line x = 4.
goo.gl/photos/v5qJLDztqsZpHR9d7
I'm just having trouble trying to set up the problem for finding the definite integral to compute the volume. But once I have it set up, I can find the solution pretty easily.
Now for b, since it's revolving about the y-axis, the definite integral setup was given as V = 4pi ∫[0,4] x(4-x)sqrt(x) dx. How did they get that formula exactly and I thought it was supposed to be 2pi not 4pi?
I think I understand how they got the "(4-x)sqrt(x)" part since when solving for y by itself, y = +- sqrt(x)(4-x), but where did that extra x in that definite integral come from???
3 answers
Thanks!
Would we want to use shells of thickness dx and symmetry for (c) as well?
Would we want to use shells of thickness dx and symmetry for (c) as well?
nope - washers, since it's revolving around a horizontal axis, just as in (a).