What is the volume generated about the x-axis but this curve?

Question

$y=-0.1x + 1$ Limits are 0 to 5 When squaring y do we expand it as $a^2 +b^2 +2ab$ and then integrate, or increase the power of the brackets to 3 and divide by 3 and the differential inside the brackets? Please explain

Answer 1

Why settle for one when you can have both? $$\begin{align}\int_0^5\left(1-\frac x{10}\right)^2dx&=\left.-\frac{10}3\left(1-\frac x{10}\right)^3\right|_0^5=-\frac{10}3\left(\frac18-1\right)=\frac{35}{12}\\ &=\int_0^5\left(1-\frac x5+\frac{x^2}{100}\right)dx=\left.\left(x-\frac{x^2}{10}+\frac{x^3}{300}\right)\right|_0^5=\frac{35}{12}\end{align}$$ Multiply by $\pi$ as desired to get the volume of the solid of revolution.