Volume of solid $y = x−4x^2$ revolved about y-axis using shell approach.

I am attempting to solve the following problem and would like some validation in my approach/need some help on finding zeros if this is indeed the correct approach.

Problem:

Find vol of solid of revolution - The region bounded by $y = x−4x^2$ and the $x$-axis revolved about the $y$-axis.

My approach:

1. Shells
2. $V= 2\pi$ * [integral of $\int_a^b x(x-4x^2)dx$
3. evaluate from b to a, and I'm assuming answer would be in pi cubic units because we're solving for volume.

How would I determine the bounds, and is my approach the correct one? (apologies for the poor formatting, I am new to the site)

Thanks, J

• Typically to find bounds, I am given two equations, which i can set to zero and solve for, which is why this problem is throwing me off. Jul 20 '18 at 4:23
• What is the appropriate range of values for $x$ (where does the parabola intersect the $x$-axis)? Jul 20 '18 at 4:44

First of all bounds $x-4x^2=0$ has roots $x=0, 0.25$. So these are your bounds. You have the integral setup correctly. So just evaluate the following. $$\int_0^{0.25}2\pi x(x-4x^2)dx$$
Because you have $f(x)=x(1-4x)$ as the top curve and $y=0$ i.e the x-axis as the bottom curve. You need to find the intersection of these curves to get the bounds.1