Finding the volume of a solid by rotating two curves about the $y$-axis

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the $y$-axis. $$y=56x−7x^2 \mbox { and }y=0$$

My issue with this question is that I am having trouble turning the equation, $y=56x−7x^2$, in terms of $y$. I understand that by doing that, I can proceed with the integration... Perhaps there is another method to do this without having to turn it in terms of $y$?

Thanks

Answer 1

Divide everything by $7$, giving

$$\frac{y}{7} = 8x - x^2.$$

Multiply by $-1$ and complete the square:

$$16 - \frac{y}{7} = (x-4)^2, \text{therefore } x = 4 \pm \sqrt{16 - \frac{y}{7}}.$$

Let me know if you have problems taking from here. Best wishes.

Answer 2

Following the advice from John Habert we can use the cylindrical shell method.

$V = 2 \pi \int_{a}^{b} x f(x) dx $

It is very helpful to draw the picture and a representative cylindrical shell

To get the x limits we solve $56x - 7x^2 = 0$

Where lower x limit , $a = 0$

Upper x limit , $b = 8$

Height of a representative shell . $f(x) = 56x - 7x^2$

$V = 2 \pi \int_{0}^{8} x(56x - 7x^2) dx$

Simplify , then integrate.