You want to find the volume of the solid obtained by rotating about the x-axis the region under the curve y = x 1 / 6 y=x1/6 from 0 to 4. you first slice through the rotated solid at a generic point x x and get a circular cross-section. what is the area of the circular cross-section?

Area = πr², where "r" is some distance "y" and/or the function "(1/6)x"; depending on the situation

Step-by-step explanation:

If I'm picturing this correctly, you'll have conical shape after revolving the function about the x-axis. If you took some generic slice and wanted to find the area of the resulting cross-section, then you would have a circle whose radius is some arbitrary value of the line that matches the slice.

For example:

y = (1/6)x right?

If you took a slice at x = 2, then the radius of the resulting cross-sectional circle would be equal to y = (1/6)•2 =1/3.

From here you just plug it into the area of a circle, πr², to get an area of π/3.

Except with an integral you need to take all the points on the interval, so the radius comes out to be the function itself.

Assuming your integral is in terms of dx, r=y. But in order to integrate in terms of dx you must replace "y" with its function (1/6)x. So ultimately r=(1/6)x and Area = π(1/6)x.

found it xd it ok yt