Find the volume of the solid that results when the region enclosed by y = x^2 and y = x^3 is revolved about the line x = 1.

Answer 1

The area "enclosed by the curves is from x = 0 to x=1 with x² being larger than x³ at all points.

You can use shell or washer method for generating the volume of revolution.

Shell seems easier (washer requires x(y) and integrating in y: integral of 2πrh(r)dr where r is the distance to the axis or |x-1| = 1-x , and h is x²-x³

V = integral from x = 0 to 1 of 2π(1-x) (x²-x³) dx

V = integral of 2π(x²-2x³+x⁴) from x = 0 to 1 = 2π(1/3-(1/2)+1/5)= π/15

Please consider a tutor. Take care.

Answer 2

A sketch really helps!!

The volume is the area of washer cross section integrated between y = 0 and y = 1. Determine the area of the cross section by finding the area of the outside circle (where y = x²) and subtracting the area of the inside circle (where y = x³). You'll have to find the radius in terms of "y". Example, the radius of the outside circle is 1 minus the "x-value" but, since y = x² then x = √y so the radius is "1 - √y"

Does this help?