Math 8 · 8.G.C.9 · G.GMD.A.2

Volume of a sphere

A ball has no flat face to stand on and no straight edge to measure. How can its volume be found from its radius alone?

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SpheresMath 8

Where you are 7 in Volume of solids

Geometry11

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Volume of solids7

Volume of a sphere

Volume of prisms and cylinders Cavalieri's principle Volume of a cylinder Volume of pyramids and cones ▾ Show all 7 (2 more)

When you reach for it

A sphere is every point at one fixed distance, the radius, from a center (see Spheres). The solid it encloses is a ball, and “the volume of a sphere” always means the volume of that ball: how much a round tank holds, how much rubber or metal a ball contains, how much air fills a balloon.

A ball is harder to measure than a box or a can. A box is so many layers of unit cubes, and a can is the same circle stacked through its height (Volume of a cylinder). A ball’s slices are circles too, but they grow from nothing at the bottom to the widest circle at the middle and shrink back to nothing at the top. Yet the answer turns out to need only the radius, and it is tied exactly to the cylinder around the ball.

Reach for this skill whenever the solid is round in every direction: a ball, a bubble, a planet treated as a ball, a dome or a bowl that is half a ball, a hollow ball such as a shell, or a liquid level that rises when balls are dropped in. It rests on the volume of a cylinder, \pi r^2 h, and on Cube roots and higher roots when the formula is run backwards.

This lesson takes two days. Part 1 gives the formula, where it comes from, and how to use it from a radius, a diameter or a circumference, for whole balls and half balls, in any unit. Part 2 runs the formula backwards to find a radius, compares balls of different sizes, and measures hollow balls and the water that balls push up. Examples 1 to 4 and Exercises 1 to 12 belong to Part 1; Examples 5 to 9 and Exercises 13 to 25 belong to Part 2.

The procedure

Part 1: the formula and how to use it

Two thirds of the cylinder around it. Put a ball of radius r inside the cylinder that just contains it, touching the bottom, the top and the side all round. That cylinder has the ball’s radius r, and its height is the ball’s full width, 2r. Its volume is

\pi r^2 \cdot 2r = 2\pi r^3

More than two thousand years ago Archimedes proved that the ball fills exactly two thirds of this cylinder. Two thirds of 2\pi r^3 is \displaystyle \frac{4}{3}\pi r^3, so the volume of a ball of radius r is

V = \frac{4}{3}\pi r^3

Figure 1The cylinder that just contains a ball of radius r has radius r and height 2r. The ball fills two thirds of it.

Why two thirds. Cut the ball through its center into two hemispheres, half balls, and stand one on its flat face. A flat cut at height y above the floor meets it in a circle. A point on the edge of that circle is r from the center of the flat face and y above it, so by the Pythagorean theorem the circle’s radius squared is r^2 - y^2, and its area is

\pi(r^2 - y^2) = \pi r^2 - \pi y^2

That is the area of a circle of radius r with a circle of radius y taken out of it: exactly the slice, at the same height, of a cylinder of radius r and height r from which a cone has been hollowed, point down. The slices match at every height, so by Cavalieri’s principle the two solids have the same volume. The hemisphere is the cylinder minus the cone, \displaystyle \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3, and the ball is two of them. The full argument, with its figure, is in Why the volume formulas hold. The same fact can be seen with water: fill a hemisphere of radius r and pour it into a cylinder of radius r, and the water stands two thirds of the way up, at height \displaystyle \frac{2}{3}r.

Find the radius first. The formula needs the radius. A diameter, the full width through the center, is twice the radius, so halve it. The widest circle round a ball, a great circle, has the ball’s center as its own center, so a tape measure round the middle gives a circumference C with C = 2\pi r (Circumference), and \displaystyle r = \frac{C}{2\pi}. The radius is cubed, so using the diameter in its place would give (2r)^3 = 8r^3: a volume eight times too large.

Cube the radius, then multiply. In \displaystyle \frac{4}{3}\pi r^3 the exponent belongs to r alone, by the order of operations: work out r^3 = r \cdot r \cdot r, then multiply by \displaystyle \frac{4}{3} and by \pi, in either order.

A hemisphere is half. A plane through the center cuts the ball into two congruent halves, so a hemisphere has volume

V = \frac{2}{3}\pi r^3

This holds only for a cut through the center. A slice taken off nearer the top is a smaller piece, and its volume is not a simple fraction of the ball’s.

Keep one unit; exact first, rounded last. The radius is multiplied by itself three times, so a radius in centimeters gives a volume in cubic centimeters. The exact volume is a number times \pi, such as 36\pi\ \text{cm}^3; round only at the end. Capacities convert as in Volume and capacity: 1\ \text{cm}^3 = 1\ \text{mL}, 1000\ \text{cm}^3 = 1\ \text{L}, and 1\ \text{m}^3 = 1000\ \text{L}.

Part 2: running the formula backwards, and comparing

Finding the radius. If V is known, \displaystyle V = \frac{4}{3}\pi r^3 is an equation in r alone. Multiply both sides by 3 and divide by 4\pi to leave r^3 by itself, then take the cube root:

r^3 = \frac{3V}{4\pi}, \qquad r = \sqrt[3]{\frac{3V}{4\pi}}

Unlike a square root, a cube root brings no second answer to set aside: every real number has exactly one real cube root, and a positive volume gives a positive radius.

Comparing sizes. Multiply the radius by k and the new volume is

\frac{4}{3}\pi (kr)^3 = k^3 \cdot \frac{4}{3}\pi r^3

so the volume is multiplied by k^3. Doubling the radius multiplies the volume by 8; a ball 10\% wider holds 1.1^3 \approx 1.33 times as much. Diameters scale by the same k, so the ratio of two volumes is also the cube of the ratio of the diameters. Run backwards, a ball that must hold m times as much needs its radius multiplied by \displaystyle \sqrt[3]{m}.

Hollow balls. A hollow ball, such as a shell or a rubber ball with air inside, is a ball of outer radius R with a ball of inner radius r taken out, both with the same center. The volume of material is the whole minus the hole:

V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3)

Figure 2A hollow ball cut through its center. The material is the ball of radius R with the ball of radius r taken out.

Water pushed up. A ball lowered into water until it is completely covered pushes the water up by its own volume, since it takes up room the water used to fill. In an upright cylindrical container of radius a, a rise of d is a cylinder of water of volume \pi a^2 d, so setting that equal to the ball’s volume gives the rise.

Worked examples

Example 1

A volume from the radius

A ball has radius 7 cm. Find its volume exactly, and to one decimal place.

Solution

Cube the radius first: 7^3 = 343. Then

V = \frac{4}{3}\pi \cdot 343 = \frac{1372\pi}{3}\ \text{cm}^3

That is the exact answer. With a calculator’s value of \pi,

V = \frac{1372\pi}{3} \approx 1436.8\ \text{cm}^3

Check the size: \pi is a little more than 3, so the volume is a little more than \displaystyle \frac{4}{3} \cdot 3 \cdot 343 = 1372, and 1436.8 is.

Example 2

A volume from the diameter

A tennis ball is 6.6 cm across. Treating it as a solid ball, find its volume to one decimal place.

Solution

“Across” is the diameter, so the radius is half of it, r = 3.3 cm, and 3.3^3 = 35.937. Then

\begin{aligned} V &= \frac{4}{3}\pi \cdot 35.937 \\ &= 47.916\pi \approx 150.5\ \text{cm}^3 \end{aligned}

Had 6.6 gone into the formula as the radius, the cube would have been 6.6^3 = 8 \cdot 3.3^3, and the volume 8 \cdot 47.916\pi \approx 1204.3\ \text{cm}^3, eight times too large. A square makes the diameter’s mistake four times too large, as with the cylinder; a cube makes it eight.

Example 3

A volume from the circumference

A tape measure round the widest part of a ball reads 75 cm. Find the ball’s volume to the nearest cubic centimeter, and say roughly how many liters that is.

Solution

The tape measures a great circle, whose radius is the ball’s radius. From C = 2\pi r,

r = \frac{75}{2\pi}\ \text{cm}

Keep that exact. Cubing it cubes the top and the bottom, \displaystyle r^3 = \frac{421875}{8\pi^3}, and then

\begin{aligned} V &= \frac{4}{3}\pi \cdot \frac{421875}{8\pi^3} \\ &= \frac{140625}{2\pi^2} \approx 7124\ \text{cm}^3 \end{aligned}

because one factor of \pi on top cancels one of the three below, leaving \pi^2. Since 1000\ \text{cm}^3 is a liter, the ball holds about 7.1 liters.

Rounding the radius first to r \approx 11.9 cm gives \displaystyle \frac{4}{3}\pi \cdot 11.9^3 \approx 7059\ \text{cm}^3, about 65\ \text{cm}^3 short: the small rounding in r is cubed and then multiplied by \displaystyle \frac{4}{3}\pi.

Example 4

A hemispherical tank in liters

A water tank is a hemisphere, flat side down, with inside radius 1.2 m. How many liters does it hold when full?

Solution

A hemisphere is half a ball, so its volume is \displaystyle \frac{2}{3}\pi r^3. With r = 1.2 m, r^3 = 1.728, and

\begin{aligned} V &= \frac{2}{3}\pi \cdot 1.728 \\ &= 1.152\pi\ \text{m}^3 \approx 3.62\ \text{m}^3 \end{aligned}

Each cubic meter is 1000 liters, so the tank holds 1152\pi \approx 3619 liters.

Check in centimeters: r = 120 cm and 120^3 = 1\,728\,000, so \displaystyle V = \frac{2}{3}\pi \cdot 1\,728\,000 = 1\,152\,000\pi\ \text{cm}^3, and dividing by 1000\ \text{cm}^3 per liter gives 1152\pi liters again.

Example 5

Finding the radius

A ball is to hold 500\ \text{cm}^3. What radius does it need, to two decimal places, and how wide is it?

Solution

Put the volume into the formula. The unknown is the radius:

500 = \frac{4}{3}\pi r^3

Multiply both sides by 3 and divide by 4\pi, so that r^3 stands alone:

r^3 = \frac{1500}{4\pi} = \frac{375}{\pi} \approx 119.37

The cube root undoes the cube:

r = \sqrt[3]{\frac{375}{\pi}} \approx 4.92\ \text{cm}

The ball is about 2 \times 4.924 \approx 9.85 cm across. There is no negative answer to set aside, since 119.37 has only one real cube root.

Check: \displaystyle \frac{4}{3}\pi \cdot 4.92^3 \approx 498.9, just under 500 because the radius was rounded down.

Example 6

Comparing balls

(a) A ball’s radius is doubled. What happens to its volume?

(b) One orange is 10 cm across and another is 8 cm across. Treating both as balls, how many times as much does the larger hold?

Solution

(a) The radius is multiplied by k = 2, so the volume is multiplied by k^3 = 8. Directly: \displaystyle \frac{4}{3}\pi (2r)^3 = \frac{4}{3}\pi \cdot 8r^3, eight times \displaystyle \frac{4}{3}\pi r^3. Each of the three factors of r in r^3 doubles.

(b) The ratio of the diameters is \displaystyle \frac{10}{8} = \frac{5}{4}, and the radii are in the same ratio, so the volumes are in the ratio

\left(\frac{5}{4}\right)^3 = \frac{125}{64} \approx 1.95

The larger orange is only 25\% wider but holds almost twice as much.

Example 7

A tank that holds twice as much

A ball-shaped gas tank has inside radius 2 m. A new tank of the same shape must hold twice as much. What radius does it need, to two decimal places?

Solution

Multiplying the radius by k multiplies the volume by k^3, and the volume is to be multiplied by 2. So

k^3 = 2, \qquad k = \sqrt[3]{2} \approx 1.26

and the new radius is \displaystyle 2\sqrt[3]{2} \approx 2.52 m.

Check with the formula itself. The old tank holds \displaystyle \frac{4}{3}\pi \cdot 2^3 = \frac{32\pi}{3}\ \text{m}^3, so the new one must hold \displaystyle \frac{64\pi}{3}. Then

r^3 = \frac{3}{4\pi} \cdot \frac{64\pi}{3} = 16, \qquad r = \sqrt[3]{16} \approx 2.52\ \text{m}

the same, since \displaystyle \sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}. The new tank is only about 26\% wider than the old one. A tank twice as wide would hold eight times as much.

Example 8

A hollow rubber ball

A hollow rubber ball has outer radius 3.5 cm, and its rubber wall is 0.5 cm thick. Find the volume of rubber to one decimal place. The rubber has density 1.1\ \text{g/cm}^3; find the ball’s mass, ignoring the air inside.

Solution

The wall runs inward from the outside surface, so the inner radius is r = 3.5 - 0.5 = 3 cm, and the outer radius is R = 3.5 cm. The rubber is the outer ball minus the inner one:

\begin{aligned} V &= \frac{4}{3}\pi (R^3 - r^3) \\ &= \frac{4}{3}\pi (42.875 - 27) \\ &= \frac{127\pi}{6} \approx 66.5\ \text{cm}^3 \end{aligned}

(Here \displaystyle \frac{4}{3} \cdot 15.875 = \frac{63.5}{3} = \frac{127}{6}.) Mass is density times volume, so the ball has mass \displaystyle 1.1 \times \frac{127\pi}{6} \approx 73.1 g.

The wall is only a seventh of the radius, yet the rubber is \displaystyle \frac{R^3 - r^3}{R^3} = \frac{15.875}{42.875} = \frac{127}{343} \approx 0.37 of the whole ball’s volume, more than a third. The outer layers of a ball are its widest, so much of a ball’s volume lies close to its surface.

Example 9

Marbles in a vase

A cylindrical vase has inside radius 4 cm. It holds water 10 cm deep, 2.1 cm below the brim. Glass marbles of radius 1 cm are dropped in one at a time and sink. How many can go in before the water spills over?

Solution

The marbles are 2 cm across and the water is 10 cm deep, so even piled two or three deep on the bottom they stay completely covered, and each pushes up its whole volume. The room left above the water is a cylinder of radius 4 cm and height 2.1 cm:

\pi \cdot 4^2 \cdot 2.1 = 33.6\pi\ \text{cm}^3

One marble has volume \displaystyle \frac{4}{3}\pi \cdot 1^3 = \frac{4}{3}\pi\ \text{cm}^3. The factors of \pi cancel in the division:

33.6\pi \div \frac{4}{3}\pi = 33.6 \cdot \frac{3}{4} = 25.2

So 25 marbles fit. Together they push up \displaystyle 25 \cdot \frac{4}{3}\pi = \frac{100\pi}{3}\ \text{cm}^3, which raises the water by \displaystyle \frac{100\pi}{3} \div 16\pi = \frac{25}{12} \approx 2.08 cm, just short of the brim. A 26th marble would bring the rise to \displaystyle \frac{26}{12} \approx 2.17 cm, more than the 2.1 cm of room, and the water would spill.

Boundaries

The formula is for a ball, round in every direction. An egg, a rugby ball or a squashed ball has slices of the wrong sizes, and \displaystyle \frac{4}{3}\pi r^3 does not apply to it as it stands.

A sphere itself has no volume. The sphere is the surface, which has an area (Surface area of a sphere); the volume belongs to the ball inside it.

r is the radius. A diameter or a circumference has to be turned into a radius first, and a mistake there is cubed.

Half a ball only through the center. The hemisphere is half because the cut passes through the center. A piece cut off by any other plane is not half, and it is not its share of the height either: the top quarter of a ball’s height holds well under a quarter of its volume.

Covered completely. A ball pushes water up by its whole volume only if it is entirely under water; a ball floating or poking out pushes up only its submerged part.

One unit, and decimals are approximations. The three lengths multiplied in r^3 share one unit, and the answer is in that unit cubed. Any decimal volume is rounded; the exact volume is a number times \pi, or, as in Example 3, a number divided by a power of \pi.

A real container has walls. What a round tank holds is measured with its inside radius; the material it is made of is a hollow ball, outside minus inside.

Where this goes next

The surface of a ball has area 4\pi r^2, and the two formulas are tied together: a ball is like a great many thin pyramids with their tips at the center and their bases covering the surface, so \displaystyle V = \frac{1}{3} \cdot 4\pi r^2 \cdot r, which is \displaystyle \frac{4}{3}\pi r^3 (Surface area of a sphere). A grain silo topped with a dome, a capsule-shaped tank or an ice cream cone with a scoop on top is split into cylinders, cones and half balls and added up (Composite solids and density), and treating a planet, a drop or a fruit as a ball is part of Modeling with solids: density and design.

Spinning a half disc about its straight edge sweeps out a ball, one of the Solids of revolution. In calculus the slice areas \pi(r^2 - y^2) used above are added up exactly (Volumes with known cross sections), and the same slicing finds the volume of a piece cut off a ball by any plane. And when a balloon is blown up at a steady rate, its radius grows more and more slowly, because the volume grows with the cube of the radius: a question in related rates.

Exercises

Part 1: finding the volume

Give each volume exactly, in terms of \pi, and to one decimal place unless asked otherwise.

A ball has radius 10 cm.
A ball has radius 1.2 feet.
A ball has radius 0.3 m. Give the decimal to two decimal places.
A ball has diameter 28 cm.
A ball has diameter 1 inch. Give the decimal to two decimal places.
A hemisphere has radius 15 cm.
A ball’s great circle has circumference 26\pi cm.

Part 1: units and capacity

A round water tank is a ball of inside radius 1.8 m. Find its volume in cubic meters, to two decimal places, and its capacity to the nearest liter.
A steel ball bearing is 9 mm across. Find its volume in cubic millimeters, to one decimal place, and in cubic centimeters, to two decimal places.
A round glass fishbowl is close to a ball of inside diameter 25 cm. How many liters does it hold when filled to the brim, to one decimal place?

Part 1: reasoning

Using the cylinder that just contains a ball, and the fact that the ball fills two thirds of it, explain where the \displaystyle \frac{4}{3} and the r^3 in \displaystyle V = \frac{4}{3}\pi r^3 come from.
Which holds more: a ball of radius 2.5 cm or a cube with edges 4 cm long? Find both volumes.

Part 2: working backwards

A ball has volume \displaystyle \frac{32000\pi}{3}\ \text{cm}^3. Find its radius.
A ball has volume 800\ \text{cm}^3. Find its radius, exactly and to two decimal places.
A hemispherical dome encloses 3888\pi\ \text{m}^3. Find its radius.
A ball holds 36 liters. Find its radius exactly, and its diameter to one decimal place.

Part 2: scaling and comparing

A ball’s radius is tripled. What happens to its volume?
A beach ball is 45 cm across and a handball is 18 cm across. How many times as much air does the beach ball hold, to one decimal place?
A ball of clay of radius 6 cm is rolled into small balls of radius 1.2 cm, with no clay left over. How many small balls are made?
A ball fits exactly inside a cube-shaped box with edges s long, touching all six faces. What fraction of the box does the ball fill, exactly and to three decimal places?

Part 2: hollow balls, water and combined solids

A hollow metal ball has outer radius 10 cm and a wall 1 cm thick. Find the volume of metal, to one decimal place.
A solid ball is dropped into a cylindrical tank of inside radius 15 cm, holding water deep enough to cover it. It sinks, and the water level rises by 2.5 cm. Find the ball’s radius.
An ice cream cone is a cone of radius 2.7 cm and height 10 cm, filled level with ice cream and topped with a half ball of ice cream of the same radius. Find the volume of ice cream, to one decimal place.

Part 2: reasoning

Explain, from the formula, why doubling a ball’s radius multiplies its volume by 8, while doubling a cylinder’s height multiplies its volume only by 2.
A cone and a ball have the same radius r. How tall must the cone be to hold the same volume as the ball? Explain.

Answers

\displaystyle \frac{4000\pi}{3} \approx 4188.8\ \text{cm}^3. 10^3 = 1000 and \displaystyle \frac{4}{3}\pi \cdot 1000 = \frac{4000\pi}{3}.
2.304\pi \approx 7.2\ \text{ft}^3. 1.2^3 = 1.728 and \displaystyle \frac{4}{3} \cdot 1.728 = 2.304.
0.036\pi \approx 0.11\ \text{m}^3. 0.3^3 = 0.027 and \displaystyle \frac{4}{3} \cdot 0.027 = 0.036.
\displaystyle \frac{10976\pi}{3} \approx 11494.0\ \text{cm}^3. The radius is 14 cm, 14^3 = 2744, and \displaystyle \frac{4}{3}\pi \cdot 2744 = \frac{10976\pi}{3}.
\displaystyle \frac{\pi}{6} \approx 0.52\ \text{in}^3. The radius is \displaystyle \frac{1}{2} inch, so \displaystyle \frac{4}{3}\pi \cdot \frac{1}{8} = \frac{\pi}{6}.
2250\pi \approx 7068.6\ \text{cm}^3. \displaystyle \frac{2}{3}\pi \cdot 15^3 = \frac{2}{3}\pi \cdot 3375 = 2250\pi.
\displaystyle \frac{8788\pi}{3} \approx 9202.8\ \text{cm}^3. 2\pi r = 26\pi gives r = 13, and \displaystyle \frac{4}{3}\pi \cdot 2197 = \frac{8788\pi}{3}.
7.776\pi \approx 24.43\ \text{m}^3, which is 7776\pi \approx 24429 liters. 1.8^3 = 5.832 and \displaystyle \frac{4}{3} \cdot 5.832 = 7.776; each cubic meter is 1000 liters.
121.5\pi \approx 381.7\ \text{mm}^3, or about 0.38\ \text{cm}^3. The radius is 4.5 mm, so \displaystyle \frac{4}{3}\pi \cdot 91.125 = 121.5\pi; a cubic centimeter is 10 \times 10 \times 10 = 1000 cubic millimeters.
About 8.2 liters. The radius is 12.5 cm, so the volume is \displaystyle \frac{4}{3}\pi \cdot 1953.125 = \frac{15625\pi}{6} \approx 8181.2\ \text{cm}^3, and 1000\ \text{cm}^3 is a liter.
The cylinder that just contains the ball has radius r and height 2r, so its volume is \pi r^2 \cdot 2r = 2\pi r^3: the r^3 is r^2 from the base times r from the height, because the height is itself twice the radius. Two thirds of 2\pi r^3 is \displaystyle \frac{2}{3} \cdot 2\pi r^3 = \frac{4}{3}\pi r^3, so the \displaystyle \frac{4}{3} is two thirds of 2.
The ball holds more: \displaystyle \frac{125\pi}{6} \approx 65.4\ \text{cm}^3 against 64\ \text{cm}^3 for the cube, since 2.5^3 = 15.625, \displaystyle \frac{4}{3} \cdot 15.625 = \frac{125}{6}, and 4^3 = 64. The ball is 5 cm across, so it would not fit inside the cube.
r = 20 cm. \displaystyle r^3 = \frac{3V}{4\pi} = \frac{32000\pi}{4\pi} = 8000, and \displaystyle \sqrt[3]{8000} = 20.
\displaystyle r = \sqrt[3]{\frac{600}{\pi}} \approx 5.76 cm. \displaystyle r^3 = \frac{3 \cdot 800}{4\pi} = \frac{600}{\pi}, then take the cube root.
r = 18 m. \displaystyle \frac{2}{3}\pi r^3 = 3888\pi gives r^3 = 5832, and \displaystyle \sqrt[3]{5832} = 18.
\displaystyle r = \frac{30}{\sqrt[3]{\pi}} cm, and the diameter is about 41.0 cm. 36 liters is 36000\ \text{cm}^3, so \displaystyle r^3 = \frac{3 \cdot 36000}{4\pi} = \frac{27000}{\pi}, and \displaystyle \sqrt[3]{27000} = 30.
It is multiplied by 27. Here k = 3 and k^3 = 27.
About 15.6 times as much. The ratio of diameters is \displaystyle \frac{45}{18} = \frac{5}{2}, so the ratio of volumes is \displaystyle \left(\frac{5}{2}\right)^3 = \frac{125}{8}.
125 balls. The radius ratio is \displaystyle \frac{6}{1.2} = 5, so the big ball holds 5^3 = 125 times as much clay as one small ball.
\displaystyle \frac{\pi}{6} \approx 0.524 of the box. The ball has radius \displaystyle \frac{s}{2}, so its volume is \displaystyle \frac{4}{3}\pi \cdot \frac{s^3}{8} = \frac{\pi s^3}{6}, and the box holds s^3.
\displaystyle \frac{1084\pi}{3} \approx 1135.2\ \text{cm}^3. The inner radius is 9 cm, so the metal is \displaystyle \frac{4}{3}\pi(10^3 - 9^3) = \frac{4}{3}\pi \cdot 271.
r = 7.5 cm. The water pushed up is \pi \cdot 15^2 \cdot 2.5 = 562.5\pi\ \text{cm}^3, so \displaystyle \frac{4}{3}\pi r^3 = 562.5\pi, r^3 = 421.875, and \displaystyle \sqrt[3]{421.875} = 7.5.
37.422\pi \approx 117.6\ \text{cm}^3. The cone holds \displaystyle \frac{1}{3}\pi \cdot 2.7^2 \cdot 10 = 24.3\pi and the half ball \displaystyle \frac{2}{3}\pi \cdot 2.7^3 = \frac{2}{3}\pi \cdot 19.683 = 13.122\pi.
In \displaystyle \frac{4}{3}\pi r^3 the radius appears as a factor three times, in r^3 = r \cdot r \cdot r. Doubling r doubles all three factors, (2r)^3 = 8r^3, so the volume is multiplied by 2 \cdot 2 \cdot 2 = 8. In \pi r^2 h the height is a single factor, so doubling it doubles the volume and no more.
Its height must be 4r, twice the ball’s diameter. Setting the volumes equal, \displaystyle \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3, and dividing both sides by \displaystyle \frac{1}{3}\pi r^2 (which is not zero) gives h = 4r. A cone holds a third of its cylinder, and a ball two thirds of a cylinder of height 2r, so the cone needs twice that height again to catch up.