11.4. Volume of a Sphere

Now, let us see how to go about measuring the volume of a sphere. First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container. Then, fill the container up to the brim with water. Now, carefully place one of the spheres in the container. Some of the water from the container will overflow into the trough in which it is kept. Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water overflowed.

To measure the volume of a sphere, we can use water displacement. By filling a container with water to the brim and then immersing a sphere in it, we can observe how much water overflows into a trough. The volume of water displaced is equal to the volume of the sphere. Therefore, if we know the radius of the sphere, we can calculate its theoretical volume using the formula $\frac{4}{3} \pi r^3$, where $r$ is the radius.

Imagine you have a balloon filled with water that has a specific shape. If you put this balloon into a bowl of water, some water will spill out. By measuring the amount of spilled water, you can find out the volume of the balloon, just like we find the volume of the sphere using the concept of displacement.

Suppose the radius of the immersed sphere is r (you can find the radius by measuring the diameter of the sphere). Then evaluate 4/3 πr³. Do you find this value almost equal to the measure of the volume over flowed?

The formula to calculate the volume of a sphere is $\frac{4}{3} \pi r^3$. When we measure the radius (by diving the diameter by 2), we can input that value into the formula to compute the volume of the sphere. This volume will be approximately equal to the volume of water displaced when the sphere is submerged, confirming that our theoretical volume matches our experimental observations.

Think of a solid ball-like a basketball or soccer ball. When you place it under a tap and it fills up with water, you can later measure how much water is in the container using a measuring cup. The water that spilled out when you placed the ball in originally represents the volume of the ball itself.

Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is $\frac{2}{3} \pi r^3$.

A hemisphere is simply half of a sphere. Therefore, to find the volume of a hemisphere, we can take the previously mentioned formula for a sphere and divide it by two. This gives us the formula for the volume of a hemisphere as $\frac{2}{3} \pi r^3$. This means that the volume of half a sphere is derived directly from the full sphere's volume, allowing for an easy calculation when dealing with hemispherical shapes.

Imagine a water bottle that is shaped like a sphere, and you cut it in half to create two hemispherical cups. Each cup would hold exactly half the amount of water that the original sphere could hold. So if the full sphere is like the bottle filled with water, each half-cup can hold half of that water.

Let's take some examples to illustrate the use of these formulae. Example 10: Find the volume of a sphere of radius 11.2 cm. Solution: Required volume = (4/3)πr³ = (4/3) × (22/7) × 11.2 × 11.2 × 11.2 cm³ = 5887.32 cm³.

To apply the formula for volume, we plug in the radius into the formula for a sphere. In our example, if the radius is 11.2 cm, we calculate the volume using the values provided in the formula: $\frac{4}{3} \pi r^3$. After completing the calculations, we find the final volume which quantifies the space occupied by the sphere.

Think of filling up a room with basketballs. If you were to calculate how much space those basketballs take up in total, you'd apply the volume formula for each ball – that's similar to how we calculated the volume for a sphere in this example. By calculating the volume of a few spheres, you can estimate what the total space would be if you had many of them.

Example 11: A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-putt. Solution: First, we calculate the volume using the volume formula and multiply by the density to find mass.

To find the mass of an object, we use the principle that mass is equal to volume times density. First, we determine the volume of the sphere using $\frac{4}{3} \pi r^3$. Then, we multiply this volume by the density (here, 7.8 g/cm³). The resulting figure gives us the mass of the sphere.

Imagine you have a bowling ball that you know weighs a certain amount, and you need to find out how much it weighs based on its size and the material it's made of. By understanding this relationship between volume and density, you can calculate the weight of various objects based on their shapes (like spheres) and materials.

Example 12: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain? Solution: The volume is $\frac{2}{3} \pi r^3$ which equals $89.8 cm^3$.

To find the volume of the water a hemispherical bowl can hold, we can use the volume formula for a hemisphere. By plugging in the radius of the bowl, we can determine how much water it would contain in cubic centimeters.

Think about making a dessert in a bowl. If you know the bowl is half-spherical in shape, you can figure out how much liquid dessert you can pour into it without overflowing, just by knowing the radius of the bowl. This practical application is the same calculation we did with the hemispherical bowl.

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