Disc Method and Shell Method Same Answer?

Question

I know that both disc and shell method should produce the same answer in this case, but for some reason I am getting two different answers when doing it two different ways.

Question is: Rotate the area bounded by $y=(x^2-1)^2, x=0$ and $x=1$ around the y-axis.

Using Shell Method: I get ${\pi\over3}$ using: $V=2\pi*\int_{0}^1f(x) x\ dx=2\pi*\int_{0}^1(x^2-1)^2*xdx$

Using Disc Method: I get $5\pi\over3$ using: $V=\pi*\int_{0}^1f(y)^2 dy=\pi*\int_{0}^1\sqrt{\sqrt{y}+1}^2*dy$

Any hints as to why I might have one of these incorrect would be great.

Answer 1

Subtle sign error! Note that $x^2-1\lt 0$ in the interval $(0,1)$, so from $(x^2-1)^2=y$ we should get $x^2-1=-\sqrt{y}$.

Thus when we use the disk method, we need to evaluate $$\int_0^1 \pi(1-\sqrt{y})\,dy.$$

Answer 2

Check the disk method calculation. Let us find the general solution first: $$y = \left(x^{2} - 1\right)^{2}$$ $$x^{2} = 1 \pm \sqrt{y}$$ $$x = \pm \sqrt{1 \pm \sqrt{y}}.$$

However, the points $(0, 1)$ and $(1, 0)$ must be points on the graph. Thus, the correct equation is $$\boxed{f(y) = \sqrt{1 - \sqrt{y}}}.$$