Cavalieri's Principle will be used to show that the formula for the volume of a sphere holds. For this purpose, consider a hemisphere and a right cylinder with a cone removed from its interior, each with the same radius and height.
Draw a right triangle with height x, base y, and hypotenuse r. Here, x is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, y is the radius of the cross sectional circle, and r is the radius of the hemisphere.
Last weekend, Tearrik installed a spherical water tank for his house.
The volume of the Earth is approximately 259875159532 cubic miles.
Solve the volume formula of the sphere for the radius and substitute the given volume.
To determine the minimum number of balloons needed to fill up the bucket, first solve for the volume of the bucket and the volume of each balloon. To do so, recall the formulas to find the volume of a cylinder and a sphere.
|Volume of a Cylinder||Volume of a Sphere|
Ramsha bought a standard pencil whose radius is 4 millimeters and the length, not including the eraser, is 180 millimeters. After a good sharpening, the tip turned into a 12 millimeter high cone.
|Volume of a Cone||Volume of a Cylinder||Volume of a Hemisphere|
Consider the fact that baseballs commonly have a radius of 1.43 inches. With this information, the volume of a baseball can be calculated, right? What about the surface area? How much leather is needed to make the lining of a baseball?
To answer these question, the following formula can be used.
The aim of this proof is to consider areas and volumes of known figures to approximate the ratio of the surface area of the sphere to the volume of the sphere. Suppose that a sphere is filled with n congruent pyramids. Consider that the area of the base of one of those pyramids is B.
Find the amount of leather needed to make the lining of a baseball that has a radius of 1.43 inches. Round the answer to one decimal place.
Notice that the surface area of a sphere with radius r is four times the area of a circle with the same radius as the sphere. This relationship can be roughly seen in the covering of a baseball.
In talking aboutspheres, there is a fascinating case called the
Napkin Ring Problem. Consider two solids that have spherical shapes, for example, a soccer ball and the planet Earth. Each is represented in the diagram, along with its respective diameter.
The interesting fact about these two napkin rings is that, since they have the same height, they have exactly the same volume.
Also, AC is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle AOC, an expression for it can be deducted.
|A circle is the set of all the points in a plane that are equidistant from a given point.||A sphere is the set of all points in the space that are equidistant from a given point called the center of the sphere.|
As can be seen, the two definitions are identical except for the fact that a circle is a two-dimensional figure while a sphere is three-dimensional. But this is not all. There is one more gnarly relationship. To see it, consider a circle and draw a line containing one diameter.