Set up a definite integral (but do not evaluate it!) for the
area of the surface obtained by rotating the curve y =x^3/3
, 0 ≤ x ≤ 1, about the x-axis. The
integral should contain only one variable (x or y).
I figured it out.
To set up the definite integral for the area of the surface obtained by rotating the curve y = x^3/3 about the x-axis, we can use the concept of the disk method.
When we rotate the curve about the x-axis, it generates a series of disks perpendicular to the x-axis. The radius of each disk is given by the y-coordinate of the curve (which is x^3/3), and the thickness (or height) of each disk is infinitesimally small and can be represented by dx.
To find the area of each disk, we use the formula for the area of a circle: A = πr^2. In this case, the radius is x^3/3.
Now we need to consider the range for x, which is given as 0 ≤ x ≤ 1. This means we need to integrate over this range to obtain the total area.
Putting it all together, the definite integral for the area of the surface is:
∫[0 to 1] π(x^3/3)^2 dx
Note that we squared the radius since we are dealing with the area of the surface. However, we do not evaluate this integral since we were asked only to set it up.