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.
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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???
(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
Thanks!
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).
To find the volumes of the solids generated when the loop of the graph y^2 = x(4-x)^2 is revolved about different axes, we can use the method of cylindrical shells. This involves setting up definite integrals that represent the sums of the volumes of many thin cylindrical shells, each with its own height and radius.
For part (a), when revolving about the x-axis, the shell height is y = 2sqrt(x)(4-x) (taking the positive square root since we want the volume above the x-axis) and the shell radius is x. To find the volume, we need to integrate the product of the circumference of the shell (2πr) and its height. Thus, the definite integral is:
V = ∫[(0 to 4)] 2πx(2sqrt(x)(4-x)) dx
To simplify the integral, you can distribute the factors and simplify the expression inside. This will result in V = 4π ∫[0,4] (4x^2 - 2x^3)^(1/2) dx. This is the correct formula for computing the volume when revolving about the x-axis.
For part (b), when revolving about the y-axis, the shell height is still y = 2sqrt(x)(4-x), but the shell radius is now the distance from the y-axis to the curve, which is x. Again, we need to integrate the product of the circumference (2πr) and the height. Hence, the definite integral is:
V = ∫[(0 to 4)] 2πx(2sqrt(x)(4-x)) dx
This simplifies to V = 4π ∫[0,4] x(4-x) sqrt(x) dx. Note that the additional x in the integrand comes from the substitution y = sqrt(x)(4-x), which was used to solve for x in terms of y. It allows us to express the integrand in terms of just x instead of y. So the formula for this case is indeed V = 4π ∫[0,4] x(4-x) sqrt(x) dx.
As for the use of 4π instead of 2π, it could be an error or a specific convention used in the problem you are referring to. Usually, when revolving about an axis, we use 2π times the definite integral to compute the volume.