Tutorial on Torricelli's Trumpet (Painter's Paradox)

Torricelli’s Trumpet Tutorial on Torricelli’s Trumpet (Painter’s Paradox) Introduction Torricelli’s Trumpet goes by several other names, including Gabriel’s Horn, and the Painter’s Paradox. The trumpet looks like this: Contents Abstract ................................................................................................ 1 Introduction .......................................................................................... 1 Volume Integral .................................................................................... 2 Surface Integral .................................................................................... 2 Painting the Inside ................................................................................ 4 Conclusions.......................................................................................... 5 Appendix 1: Approximation using rectangles ....................................... 5 Appendix 2: The surface integral .......................................................... 6 It is the surface of revolution of the function f (x) = 1/x, starting from x = 1 and extending out to infinity. Abstract There is actually no mystery with Torricelli’s Trumpet when presented as the Painter’s Paradox. Although the internal surface area is ‘infinite’, a finite layer of paint requires a smaller volume of paint than that contained within the horn. Leslie Green CEng MIEE 1 of 8 The ‘paradox’ is that the volume of the trumpet is finite, but the surface area is infinite. How can such a surface be understood, and painted? v1.00: October 2024 Torricelli’s Trumpet Volume Integral To calculate the volume of the trumpet we need to use a simple integral. First we can sketch the top edge of the side view of the trumpet, zoomed in to just a small section such that the curved surface looks straight. If the initial 1 in the integral had units of centimetres (cm) then the volume for a sufficiently large value of L would tend to  cm³. Surface Integral y y y+y y x y=1/x y=1/x x x-axis (of rotation) x+x x Since we are looking at the upper part of the cross section, at any point x the radius of the disc will be y = 1/x. If we move a short distance x along the x-axis, the new y value will be y+y. The volume of the grey rectangle, rotated about the x-axis, will be r2x = y2x = (x)/x². Without bothering too much about the underlying mathematics of integration, we will simply claim that if x is made sufficiently small, the approximation of the grey disc to the volume of the trumpet in that region will be adequately accurate. We will demonstrate this in Appendix 1. Since we don’t want to use the infinity symbol, , as a number, we will use some large number L as the upper limit in the integral. The volume of the trumpet is then written as: x L volume =  2  dx =   1 Leslie Green CEng MIEE  1 L 1  1   1 L  dx =   −  =    − + 1 2 x  L   x 1 x-axis (of rotation) x+x We want to find the length of the green line, which will become a surface when revolved around the x-axis. 2 length =  x +  y 2 2  y   dy  x →dx ⎯→ dx  1 +   =  x  1 +   ⎯⎯  dx  x  2 The arrow with  x → dx can be read as x getting ‘sufficiently small’. (This used to be considered as the limit as x tended to zero, but that style has become unfashionable.) Since we have y = 1 = x −1 , we can differentiate this with respect to x to get x dy dx = − x −2 = − 1 , x2 which we can substitute back into the length equation for the green element: 2 of 8 v1.00: October 2024 Torricelli’s Trumpet length = 1+ 1  dx x4 The small surface area created by rotating this element about the x-axis is: surface = 2  r  1 + 2 1  1 + 4  dx x x 1  dx = x4 The total surface is again found by integration: area =  2 1  1 + 4  dx = 2  x x L 1  1 L 1 1  1 + 4  dx x x Sadly, that integral is a bit complicated. A usual trick at this point is to realise that for positive values of x, 1 1  1+ 4 x x  1 x 1 1  1 + 4  dx  x x  Thus  L 1  1 L 1 L 1  dx x L 1  dx =  ln(x )  = ln(L ) − ln(1) = ln(L ) 1 x As L tends to infinity, ln(L) tends to a smaller infinity, and the surface area of the trumpet tends to that smaller infinity. So far we have followed the usual arguments exactly. Now we want to consider painting the inside of the trumpet, quantitatively. The more complicated integral is analysed in Appendix 2. Leslie Green CEng MIEE 3 of 8 v1.00: October 2024 Torricelli’s Trumpet For the simple approximation, the volume of paint used is approximately surface area  paint thickness. This is only approximate for large paint thicknesses because the corrected radius of revolution is smaller by around half the paint thickness. It means we have a significant over-estimate of the required paint volume at larger paint thicknesses. Painting the Inside Approximate paint volume is 2     1 Accurate paint volume is 2     1 1 1 1 1  dx = 2    ln   x   1 1  1.196   1 + 4  dx = 2    ln   x x    We consider a definite paint thickness, , such that  > 0. Notice that the blue (paint) lines come together at some point. When that happens, paint cannot get further down the trumpet. When y = , the upper and lower paint layers touch. This is the limiting case. For y = , that gives L = 1/. Hence our integrals are bounded. The surface area times the paint thickness approximates the volume of paint used. The volume of paint filling the trumpet is Leslie Green CEng MIEE    1 − 1  =  (1 −  ) L 4 of 8 v1.00: October 2024 Torricelli’s Trumpet Although the surface is accurately calculated, the paint volume is in error since we didn’t correctly account for the smaller internal radius in the horn due to the finite paint thickness. Appendix 1: Approximation using rectangles For simplicity we will just use geometry. Nevertheless, the point is made that the volume of paint required is not ‘infinite’. In fact it is a strictly decreasing function of the paint thickness for thicknesses below 0.1 units. h Conclusions The 300+ year old “paradox” that an infinite amount of paint is required to paint the infinite interior surface of the horn is seen to be spurious. We have demonstrated that making the paint layer thinner actually requires less paint, which is a common observation on finite planar surfaces. w We want the area between the blue line and the horizontal axis, the area below the curve. Instead we use the area of the grey rectangle. Rather than wh/2 we have wh, an over-estimate of 100%, which is not very good! The earlier assumption that an infinite area requires an infinite amount of paint is not correct once you realise that paint cannot pass through gaps smaller than the required paint thickness. h w Instead we use 6 rectangles. The over-estimate is now half of each of the 6 little rectangles. Each little rectangle is (w/6)(h/6). The over-estimate is therefore 6(w/6)(h/6)/2 = wh/12. With n little strips the over-estimate would be wh/(2n). The area should have been wh/2, so the fractional (per- Leslie Green CEng MIEE 5 of 8 v1.00: October 2024 Torricelli’s Trumpet d  ln(x + 1) − ln(x − 1) dx Technically, this use of rectangles (step-functions) to find the area under the curve is known as Riemann integration. Using rectangles above the curve, as we have done here, is known as the upper Riemann sum. If we used rectangles just below the curve at each point, that would be the lower Riemann sum. For sufficiently large n the two sums should converge to a single value, and that is the definition of the integral. Using the function of a function rule (aka the chain rule):  ( =   ( ) d −4 x dx  = − 4 x −5 = − Alternatively, substitute u = x , ) ( ) ln x + 1 − ln x − 1  ( ) ( If you wrote this in a calculus class you would lose marks for having omitted a constant term on the right. We are deliberately omitting such constants here in order to avoid unnecessary clutter. 6 of 8 1 1  x −1 2 x so that − 1 (x − 1) x du 1 = dx 2 x = ln(u + 1) − ln(u − 1) ) d  ln x + 1 − ln x − 1 dx 1 1 1  dx = −  4 5 x 4 x x −1 2 x −1 2 1  2  − 2 x  x − 1 = ( 4 x5 = − 2 1 1  − x +1 2 x = Some people call this the anti-derivative: d  1   = dx  x 4  1 1 − = x +1 x −1 ) ( ) = 1  ( x − 1) − ( x + 1)   2 x  ( x + 1) ( x − 1)     1 1 1  1  ln 1 + 4 + 1 − ln 1 + 4 − 1 − 2 1 + 4  4   x x x     Starting from the indefinite integral (one in which the limits are not given) on the left is very difficult. Having seen the answer on the right, however, we can try to work backwards. In effect we are asking “what function, when differentiated with respect to x, gives the expression inside the integral on the left?”. Leslie Green CEng MIEE = d ln x + 1 − ln x − 1 dx Appendix 2: The surface integral 1 1  1 + 4  dx = x x  (x − 1) − (x + 1) unit) over-estimate is [wh/(2n)] / [wh/2] = 1/n. Thus for sufficiently large n, the per-unit error is arbitrarily small. = = du d    ln(u + 1) − ln(u − 1) dx du  1 −2 = − 2 (x − 1) x 2 x u −1 1  v1.00: October 2024 Torricelli’s Trumpet Now we consider u = 1+ 1 x4 du = − dx giving    d   1 1 ln 1 + 4 + 1 − ln 1 + 4 − 1 = dx   x x    du 2 = −  2 dx u − 1 du = −  2x4 dx 1+ 1 x4  4 x5 1 x4 Verification using Wolfram Alpha (Oct 2024): 4 x5  There will be 6 terms in the required definite integral, so for simplicity we will calculate the limit values separately (neglecting the 4 multiplier for now.) 4 x  1+ 1 2 1+  du d   ln(u + 1) − ln(u − 1) dx du = 1  1 d   1+ 4  = −    2 dx  x  1 = 1 x4 The lower limit is 1, giving: 4x ( x 3 ) = − 1.0657   1 ln 1 + 4 + 1 25  − 1  10 −6  ln(2) x4 + 1 We can replace the first term in the upper limit with ln(2), resulting in a negligible error. Combining those two results:    1 1 1  d   ln 1 + 4 + 1 − ln 1 + 4 − 1 − 2 1 + 4  dx   x x x     4x 4 4 x4 + 1 = + = x 4 + 1 x3 x 4 + 1 x3 x 4 + 1 ( For x  32 2  1+ ) = 4 x4 + 1 x3 = 1 − 2  1 10−6 4 32 We can replace the third term in the upper limit with −2, resulting in a negligible error. 4 1 1+ 4 x x For x > 32 we have 5 constant terms which can be gathered together:  ln(2) − 2  − which is a scaled version of the required anti-derivative. Leslie Green CEng MIEE ( For x  25 2 = − ) ln 2 + 1 − ln 2 − 1 − 2 2 x4 + 1 7 of 8  ln( ) ( ) 2 + 1 − ln 2 − 1 − 2 2  = − 0.2412 v1.00: October 2024 Torricelli’s Trumpet The definite integral has just one remaining term, where we replace the upper limit L by 1/, then use the binomial approximation for the square root: ) (   1 − ln 1 + 4 − 1 = − ln 1 +  4 − 1 x   = − ln 1 + 12  4 +  − 1  (( − ln(  ) 1 2 4 Version History v1.00: 7 October 2024. First publication on http://lesliegreen.byethost3.com ) = ) − ln( ) + 0.6931 4 = − 4  ln( ) + 0.6931 The approximate result is:  1 1 1 1  dx = ln(x ) 1 x = ln(1  ) − ln(1) = − ln( ) The more accurate result is:  1 1 1    1  1 1 1      ln 1 + 4 + 1 − ln 1 + 4 − 1 − 2 1 + 4  4   x x x  1    0.6931 0.2412 = − ln( ) + − 4 4 = − ln( ) + 0.1130 1 1  1 + 4  dx = x x = − ln( ) + ln(exp(0.1130)) ORCHID ID: 0009-0003-5554-9148 = − ln( ) + ln(1.196) = − ln(0.8932   ) Leslie Green CEng MIEE 8 of 8 v1.00: October 2024