Single-Variable-Calculus · Unit 20 · Video 1 · Interactive Practice

Probability as a Shell Volume: Hitting a Ring on a Dart Board

IKey Formulas

FormulaMeaningWhat you need
hits in ΔA≈c e−r2 ΔA\text{hits in } \Delta A \approx c\,e^{-r^2}\,\Delta AThe model: hits per unit area follow a bell curve in the distance rr from the centreThe constant cc grows with the number of throws; its value is never needed
part=c∫r1r22πr e−r2 dr=cπ(e−r12−e−r22)\displaystyle \text{part} = c\int_{r_1}^{r_2} 2\pi r\,e^{-r^2}\,dr = c\pi\left(e^{-r_1^2} - e^{-r_2^2}\right)Hits in the ring r1<r<r2r_1 < r < r_2: a volume of revolution, by shellsCircumference 2πr2\pi r, height e−r2e^{-r^2}, thickness drdr; antiderivative −πe−r2-\pi e^{-r^2}
whole=cπ(e−02−e−∞2)=cπ\text{whole} = c\pi\left(e^{-0^2} - e^{-\infty^2}\right) = c\piEvery place a dart could land, idealised out to r=∞r = \inftyr1=0r_1 = 0, r2=∞r_2 = \infty in the part formula
P(r1<r<r2)=partwhole=e−r12−e−r22\displaystyle P(r_1 < r < r_2) = \frac{\text{part}}{\text{whole}} = e^{-r_1^2} - e^{-r_2^2}Probability of landing in the ringBoth cc and π\pi cancel

Key Insight: Hits are counted per unit area of the board, so a ring's count is height e−r2e^{-r^2} times the ring's area 2πr dr2\pi r\,dr, added up: a shell volume, not the area under the bell curve. The unknown cc multiplies part and whole alike, so it drops out of the probability.

IIVisualization 1 — The Part Is a Volume, Not an Area

A ring's count is the area under 2πr e−r22\pi r\,e^{-r^2}, not under e−r2e^{-r^2}: its circumference 2πr2\pi r multiplies the height.

IIIVisualization 2 — Shells Add Up to the Part

Thin shells 2πri e−ri2 Δr2\pi r_i\,e^{-r_i^2}\,\Delta r between r1=0.5r_1 = 0.5 and r2=1.5r_2 = 1.5 add up to π(e−0.25−e−2.25)\pi\left(e^{-0.25} - e^{-2.25}\right).

IVVisualization 3 — Part over Whole: c Cancels

Whatever the number of throws, the fraction of darts in a ring settles at e−r12−e−r22e^{-r_1^2} - e^{-r_2^2}.

💡 A dart past a wall at r=3r = 3 has probability e−9≈0.0001e^{-9} \approx 0.0001, which is why idealising the whole out to r=∞r = \infty changes nothing that matters.

VQuiz Questions

Problem 1 · The Inner Disc

Given: hits in ΔA≈c e−r2 ΔA\Delta A \approx c\,e^{-r^2}\,\Delta A. Find the probability that a dart lands within distance 11 of the centre, P(0<r<1)P(0 < r < 1).

✅ Correct! With r1=0r_1 = 0 and r2=1r_2 = 1: e0−e−1=1−e−1≈0.632e^{0} - e^{-1} = 1 - e^{-1} \approx 0.632.
❌ That is the outside. e−1e^{-1} is P(1<r<∞)=e−1−e−∞P(1 < r < \infty) = e^{-1} - e^{-\infty}, the chance of landing beyond radius 11.
❌ Part, not probability. π(1−e−1)\pi(1 - e^{-1}) is the shell volume (with cc left out); divide by the whole, π\pi. A probability can never exceed 11.
❌ That is the flat area under the bell. Hits are counted per unit area of the board, so each ring needs its circumference 2πr2\pi r: the count is ∫2πr e−r2 dr\int 2\pi r\,e^{-r^2}\,dr, then divided by the whole.
Show solution

The part is the shell volume from r1=0r_1 = 0 to r2=1r_2 = 1:

part=c∫012πr e−r2 dr=c[−πe−r2]01=cπ(1−e−1).\text{part} = c\int_0^1 2\pi r\,e^{-r^2}\,dr = c\left[-\pi e^{-r^2}\right]_0^1 = c\pi\left(1 - e^{-1}\right).

The whole is cπc\pi, so

P(0<r<1)=cπ(1−e−1)cπ=1−e−1≈0.632.P(0 < r < 1) = \frac{c\pi\left(1 - e^{-1}\right)}{c\pi} = 1 - e^{-1} \approx 0.632.

About 63%63\% of darts land within distance 11 of the centre.

Problem 2 · Disc or Ring?

Given: the same model. Compare the inner disc 0<r<0.50 < r < 0.5 with the ring 0.5<r<10.5 < r < 1 of the same width. Which is more likely to be hit?

✅ Correct! 1−e−0.25≈0.2211 - e^{-0.25} \approx 0.221 for the disc and e−0.25−e−1≈0.411e^{-0.25} - e^{-1} \approx 0.411 for the ring: the ring's larger area beats its lower density.
❌ Density is not the whole story. The density is highest at the centre, but the disc is small. The ring goes all the way around, with area about 2πr2\pi r times its width, so it collects more hits.
❌ Same width, different area. The disc has area π(0.5)2≈0.79\pi(0.5)^2 \approx 0.79; the ring has area π(12−0.52)≈2.36\pi(1^2 - 0.5^2) \approx 2.36, three times as much.
❌ Those are flat areas under e−r2e^{-r^2}. ∫00.5e−r2 dr≈0.461\int_0^{0.5} e^{-r^2}\,dr \approx 0.461 and ∫0.51e−r2 dr≈0.286\int_{0.5}^{1} e^{-r^2}\,dr \approx 0.286 count hits per unit length of rr, ignoring the factor 2πr2\pi r that measures each ring's size.
Show solution

Use P(r1<r<r2)=e−r12−e−r22P(r_1 < r < r_2) = e^{-r_1^2} - e^{-r_2^2}:

P(0<r<0.5)=e0−e−0.25=1−0.7788≈0.221,P(0 < r < 0.5) = e^{0} - e^{-0.25} = 1 - 0.7788 \approx 0.221, P(0.5<r<1)=e−0.25−e−1=0.7788−0.3679≈0.411.P(0.5 < r < 1) = e^{-0.25} - e^{-1} = 0.7788 - 0.3679 \approx 0.411.

The ring is almost twice as likely. The density e−r2e^{-r^2} is lower there, but the ring has three times the area of the disc, and the count is density times area.

Problem 3 · The Whole and the Outside

Given: hits in ΔA≈c e−r2 ΔA\Delta A \approx c\,e^{-r^2}\,\Delta A. Find the whole (all rr from 00 to ∞\infty) and the probability that a dart lands farther than 1.51.5 from the centre.

What is the whole?

What is P(r>1.5)P(r > 1.5)?

✅ Correct! The whole is cπc\pi, and the outside 1.5<r<∞1.5 < r < \infty has probability e−2.25−0≈0.105e^{-2.25} - 0 \approx 0.105.
❌ That is a flat area. ∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} is the area under the bell along a line. The whole board needs the shell volume c∫0∞2πr e−r2 drc\int_0^\infty 2\pi r\,e^{-r^2}\,dr.
❌ Check the antiderivative. ddr(−πe−r2)=2πr e−r2\frac{d}{dr}\left(-\pi e^{-r^2}\right) = 2\pi r\,e^{-r^2}, so the antiderivative of 2πr e−r22\pi r\,e^{-r^2} is −πe−r2-\pi e^{-r^2}, not −2πe−r2-2\pi e^{-r^2}.
❌ Not quite. Put r1=0r_1 = 0 and r2=∞r_2 = \infty into part=cπ(e−r12−e−r22)\text{part} = c\pi\left(e^{-r_1^2} - e^{-r_2^2}\right).
❌ That is the inside. 1−e−2.251 - e^{-2.25} is P(0<r<1.5)P(0 < r < 1.5). Farther than 1.51.5 means r1=1.5r_1 = 1.5, r2=∞r_2 = \infty.
❌ Square the radius. The exponent is −r2=−(1.5)2=−2.25-r^2 = -(1.5)^2 = -2.25, not −1.5-1.5.
❌ That is the part. Divide by the whole cπc\pi: a probability contains no cc.
❌ Not quite. Use P(r1<r<r2)=e−r12−e−r22P(r_1 < r < r_2) = e^{-r_1^2} - e^{-r_2^2} with r1=1.5r_1 = 1.5 and r2=∞r_2 = \infty.
Show solution

The whole. With r1=0r_1 = 0 and r2=∞r_2 = \infty:

whole=c∫0∞2πr e−r2 dr=c[−πe−r2]0∞=cπ(1−0)=cπ.\text{whole} = c\int_0^\infty 2\pi r\,e^{-r^2}\,dr = c\left[-\pi e^{-r^2}\right]_0^\infty = c\pi\left(1 - 0\right) = c\pi.

The outside. The part from r1=1.5r_1 = 1.5 to r2=∞r_2 = \infty is cπ(e−2.25−0)c\pi\left(e^{-2.25} - 0\right), so

P(r>1.5)=cπ e−2.25cπ=e−2.25≈0.105.P(r > 1.5) = \frac{c\pi\,e^{-2.25}}{c\pi} = e^{-2.25} \approx 0.105.

About one dart in ten lands farther than 1.51.5 from the centre.

Problem 4 · A Less Accurate Thrower

Given: a different thrower, modelled by hits in ΔA≈c e−r2/2 ΔA\Delta A \approx c\,e^{-r^2/2}\,\Delta A. Find P(0<r<2)P(0 < r < 2) by redoing part and whole with shells.

✅ Correct! Part =2πc(1−e−2)= 2\pi c\left(1 - e^{-2}\right), whole =2πc= 2\pi c, so P=1−e−2≈0.865P = 1 - e^{-2} \approx 0.865.
❌ That uses the old density. 1−e−41 - e^{-4} comes from e−r2e^{-r^2}. With e−r2/2e^{-r^2/2} the exponent at r=2r = 2 is −42=−2-\tfrac{4}{2} = -2.
❌ Part, not probability. 2π(1−e−2)2\pi(1 - e^{-2}) is the shell volume with cc left out. Divide by the whole, 2π2\pi; a probability cannot exceed 11.
❌ Check the exponent. −r2/2-r^2/2 at r=2r = 2 is −4/2=−2-4/2 = -2; the value −1-1 would come from −(r/2)2-(r/2)^2.
Show solution

Antiderivative. ddr(−2πe−r2/2)=−2π⋅(−r) e−r2/2=2πr e−r2/2\frac{d}{dr}\left(-2\pi e^{-r^2/2}\right) = -2\pi\cdot(-r)\,e^{-r^2/2} = 2\pi r\,e^{-r^2/2}.

Part.

c∫022πr e−r2/2 dr=c[−2πe−r2/2]02=2πc(1−e−2).c\int_0^2 2\pi r\,e^{-r^2/2}\,dr = c\left[-2\pi e^{-r^2/2}\right]_0^2 = 2\pi c\left(1 - e^{-2}\right).

Whole.

c∫0∞2πr e−r2/2 dr=2πc (1−0)=2πc.c\int_0^\infty 2\pi r\,e^{-r^2/2}\,dr = 2\pi c\,(1 - 0) = 2\pi c.

Probability.

P(0<r<2)=2πc(1−e−2)2πc=1−e−2≈0.865.P(0 < r < 2) = \frac{2\pi c\left(1 - e^{-2}\right)}{2\pi c} = 1 - e^{-2} \approx 0.865.

In general this model gives P(r1<r<r2)=e−r12/2−e−r22/2P(r_1 < r < r_2) = e^{-r_1^2/2} - e^{-r_2^2/2}: the same part-over-whole shape, with cc cancelling again.

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