Find the volume of the solid revolved around y=2x^2 +4x, y=0, about the x-axis

Answer 1

Limits of integration: 2x² + 4x = 0; x = - 2; x = 0.

By7 disk method volume V = π∫_-2⁰(2x² + 4x)²dx = π(4x⁵/5 + 4x⁴ + 16x³/3)_-2⁰ = π(128/5 - 64 + 128/3) = 64π/15

Answer 2

I'm assuming that x=-2 and x=0 are the bounds of this question? Otherwise, the function grows unbounded and the answer is unbounded.

The way to solve this is to first create an equation for the area of each disc generated by the revolution. Each disc has a radius of y. So the equation is:

A = π * y² = π * (2x² + 4x)²

To find the volume, we evaluate the definite integral from -2 to 0 to sum up the areas of all the discs:

∫_-2⁰ (π * (2x² + 4x)²)dx

Can you complete the final steps yourself?