Verified answer

Well, your limits are correct: y = x³ and y = 9x intersect at points (0,0) and (3,27)

So we integrate from x = 0 to x = 3

On this interval 9x > x³

We rotate about x-axis (y = 0). We use washer method, where

R = outer radius = 9x

r = inner radius = x³

V = ∫₀³ (πR² − πr²) dx

V = π ∫₀³ ((9x)² − (x³)²) dx

V = π ∫₀³ (81x² − x⁶) dx

V = π (27x³ − 1/7 x⁷) |₀³

V = π ((729 − 2187/7) − (0 − 0))

V = 2916π/6

The integral you use is for finding area of the given region,

not for finding volume of given region rotated about x-axis.

Mαthmφm

you do no longer specify what to rotate the area approximately, which of course very much impacts the respond. I worked it out utilising the shell approach and rotating it around the y-axis. the respond I have been given became 38pi/9 once you rotate it the form sounds like a cylinder with a scoop out of the exceptional. the top is set via the exceptional function y=f(x)=x^7 + 4 The radius of the cylinder is x, we are finding for the area between 0 and a million. in case you're taking a cylindrical slice of the "shell", decrease it in a million/2 and flatten it it sounds like a very skinny field with V = length(2pix) *width(dx)*top(x^7+4) Take the imperative. imperative from 0 and a million 2pix(x^7 + 4)dx = 2pi imperative from 0 to a million (x^8 + 4x) dx = 2pi (x^9/9 + 2x^2) ] evaluated from 0 to a million = 2pi (a million/9 + 2) - 2pi (0) = 38pi/9