Solid of revolution

Solids of revolution in this visualization

Why both ways give the same volume

• A small piece of the region at distance $\rho$ from the axis travels a circle of length $2\pi\rho$. A strip across the axis holds pieces at every distance from $r$ to $R$, and $\int_r^R 2\pi\rho\,d\rho = \pi(R^2-r^2)$. Unroll slice straightens each circle of the ring: they stack into a trapezoid with parallel sides $2\pi r$ and $2\pi R$, the same area but not the same shape.
• A strip along the axis holds pieces all at the same distance, so its cylinder has area $2\pi\rho h$; cut along a line and laid flat it is a $2\pi\rho \times h$ rectangle. The two methods add the same pieces in two orders: they are one double integral, taken in either order.
• The centroid (the center of mass of a uniform sheet) is at distance $\bar\rho = \frac{1}{A}\iint_D \rho\,dA$ from the axis, so $V = 2\pi\bar\rho A$ (Pappus–Guldin). Moving the axis a distance $d$ further away adds exactly $2\pi d A$.

Things not to misread

• The strip and its slice are drawn with no thickness: nothing here is a step $\Delta x$, and the volume is an exact integral, not a sum of thin pieces (finite slices are the Riemann sum's page). The faint lines on the glass only show its shape.
• A ring's area is $\pi R^2 - \pi r^2$, not $\pi(R-r)^2$. About $y = -1$ the outer radius is $f(x)+1$, not $f(x)$.
• Volumes are exact: each region's area and first moments are closed forms, and Pappus gives the volume from them. The volume added up so far is computed numerically and marked ≈. Under $y = 1/x$ the area $\ln b$ grows without bound while the volume $\pi(1 - 1/b)$ stays below $\pi$.

Controls

• Drag the strip in the view or anywhere on the graph; drag its swept edge round the axis; drag the axis, which stops at the region and catches on whole and half numbers (the value keys put it on the other side). Front looks straight at the region's plane. Presentation mode opens a full-screen view: ← → move the strip, ↑ ↓ turn it.

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