Find volume of these solids using integration

Question

1. a) The $(x>0, y< -1)$ region of the curve $y= -\frac{1}{x}$ rotated about the $y$-axis. The instructions say that one should use the formula: $V = \int 2πxf(x) dx$

I used another method and got an answer of $π \text{ units}^3$. The formula that I used was $\int π* 1/y² dy$ (and I used an upper bound of $-1$, lower bound of $-\infty$). Could you please explain how they got to the formula of $V = \int 2πxf(x) dx$, and how it can calculate the volume in my situation.

1. b) i need to do the same thing, but this time with the curve $yx² = -1$ (the region is still the same: $x>0, y< -1$).

I used my own method (which is written above) but when rearranging for $x$, I had to square root both sides, ending up with $x = \pm(-1/y)^{0.5}$

But again, I am suposed to use the formula that was given: $V = \int 2πxf(x)$ So how would I use this method to find the solutions to both of these problems, and how did they derive this formula?

Any advice would be much appreciated, I'm stressing out right now as this homework is due tomorrow morning.

Answer 1

In this case, you are supposed to setup an integral that uses a function turning radial coordinate $\rho \in [a,b]$ into height $h$. In (a), the function is $ h(\rho) = 1/\rho - 1$ on $(0,1)$, since the graph is truncated by $y = -1$. The formula $V = 2\pi \int_a^b xf(x)\,dx$ is obtained through cylindrical volume integration: $$\begin{align*} \iiint_V dV &= \int_a^b \int_0^{h(\rho)}\int_0^{2\pi} \rho\,d\phi\,dz\,d\rho \\ &= \int_a^b 2\pi \rho h(\rho)\,d\rho \\ \end{align*}$$ In this case, $(x,y)$ is replaced by $(\rho,z)$, and $f(x)$ is replaced by $h(\rho)$.

Answer 2

For every $x$ such that $ 0<x\leq 1$. There is a cylinder (I'm talking about just the surface-the boundary-of the cylinder) with radius $x$ (centred at $x=0$) that extends down from $y=0$ to $y=f(x)$. This cylinder is completely within your solid, and if we extended it any lower it would not remain in your solid. Of course, this cylinder has no volume but if we give it an infinitesimal thickness, say, $dx$, then we can sum the 'volumes' of all such cylinders and obtain the total volume of the solid.

$2\pi x$ is the circumference of a circle with radius $x$. We want the depth to be zero when $f(x)=-1$ and we want the depth to be positive (to compute volume). Therefore we adjust and let the depth be given by $|f(x)|-1$. Now each cylinder has 'volume' $2\pi x (|f(x)|-1)dx$. The total volume is then $\int_0^1 2\pi x (1/x-1)dx$.