Calculating Gabriel's Trumpet volume $\left(\frac{1}{z}\right)$

Question

I want to calculate the volume of the solid obtained by rotating the function $1/z$ about the $z$-axis for $z>1$, that is $\pi$. But I want to apply that: \begin{equation*} V = 4\int \int_{\mathcal{S}} f(x,y) \,\text{d}x\text{d}y, \end{equation*} where \begin{equation*} z = f(x,y) = \frac{1}{\sqrt{x^2+y^2}}, \end{equation*} and the region $\mathcal{S}$ is $x^2+y^2<1$ for the first quadrant.

We change $x=r\cos\theta$ and $y=r\sin\theta$. Then \begin{equation*} V = 4\int \int_{\mathcal{S}} f(x,y) \,\text{d}x\text{d}y = 4\int \int_{\mathcal{T}} f(r,\theta) \frac{\partial(x,y)}{\partial(r,\theta)}\,\text{d}r\text{d}\theta. \end{equation*} Since $f(r,\theta)=1/r$ and $\frac{\partial(x,y)}{\partial(r,\theta)}=r$ we obtain \begin{equation*} V = 4\int_{0}^{\pi/2} \text{d}\theta \int_{0}^{1} \text{d}r = 2\pi \neq \pi \end{equation*}

Why I can't apply that method? Thanks.

Answer 1

That's because $z = f(x, y)$ is the vertical distance from xy-plane to a point on the surface. You need to measure the vertical distance from plane $z = 1$.

So the integral should be,

$V = \displaystyle \iint_{x^2 + y^2 \leq 1} (f(x,y) - 1) ~ dx ~ dy$

$ \displaystyle = 4 \int_0^{\pi/2}\int_0^1 r \cdot \left(\frac 1 r - 1\right) ~ dr ~ d\theta = \pi$

Answer 2

\begin{align} z&=\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r},\\ \text{d}V&=\pi r^2(z)\,\text{d}z= \frac{\pi}{z^2}\,\text{d}z,\\ \int_1^{\infty}\pi r^2(z)\,\text{d}z&=\int_1^{\infty} \frac{\pi}{z^2}\,\text{d}z=\left[\frac{\pi}{z}\right]_{\infty}^1=\pi. \end{align}