Volume of Revolution for $y=1-x^2$ and $y=2x$

Question

There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $\int_{0}^{\sqrt2-1} \pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $\int_{0}^{\sqrt2-1} \pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?

Answer 1

Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.