Question

find the volume using the disc method
f(x)=x^3 -2 between x= 0 and x= 2

Answer 1

Assuming you mean the solid formed by revolving the region between y = f(x) = x^3 − 2 and the x-axis, for 0 ≤ x ≤ 2, about the x-axis:

Radius = |f(x)|, so the volume by the disc method is
V = π ∫_0^2 [x^3 − 2]^2 dx.

Compute the integral:
[x^3 − 2]^2 = x^6 − 4x^3 + 4,
∫(x^6 − 4x^3 + 4) dx = x^7/7 − x^4 + 4x.

Evaluate from 0 to 2:
V = π [ (2^7/7 − 2^4 + 4·2) − 0 ] = π [128/7 − 16 + 8] = π (72/7) = 72π/7.

Numerical value ≈ 32.31 (units^3).