Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by y=x^4...?

Question

... and y=64, about the x-axis.

Answer 1

The intersection point is at (2√2, 64). The volume will be
π∫(0 to 2√2) (64^2 - (x^4)^2) dx
= π∫(0 to 2√2) (4096 - x^8) dx
= π[4096x - x^9 / 9] [0 to 2√2]
= π[(8192√2 - 2^(27/2) / 9) - (0 - 0)]
= 8192π√2 (1 - 1/9)
= 65536π√2 / 9.
≈ 32352.1.

Answer 2

First you have to find the intersections of the two graphs in quadrant I. x^4 = 64 -> x = 2sqrt(2). Now use the disk method: the area is pi(fnInt((64^2 - (x^4)^2, x, 0, 2sqrt(2))), which you can either just do by hand (it's pretty easy) or plug it into your calculator.