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

calculusvolume

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

Best Answer

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. $$\int_0^{0.25}2\pi x(x-4x^2)dx$$

Why the bounds are so?

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

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