A semicircle takes its diameter as its axis of rotation, and the surface formed by rotation is called a sphere.

The geometric figure surrounded by spheres is called sphere, which is called sphere for short.

The center of the semicircle is called the center of the sphere.

The line segment connecting the center of the sphere and any point on the sphere is called the radius of the sphere.

The line segment connecting two points on the sphere and passing through the center of the sphere is called the diameter of the sphere.

Spherical characteristics

Cut a ball with a plane, the cross section is round. The cross section of the ball has the following characteristics:

The straight line connecting the center of 1 sphere and the center of section is perpendicular to the section.

2 the distance d from the center of the sphere to the cross section has the following relationship with the radius r of the center of the sphere and the radius r of the cross section: r 2 = r 2-d 2.

A circle whose sphere is cut by a plane passing through the center of the sphere is called a big circle, and a circle cut by a section not passing through the center of the sphere is called a small circle.

On the sphere, the length of the shortest connecting line between two points is the length of the bad arc between these two points through the great circle of these two points. We call this arc length the spherical distance between two points.

Spherical function

The equation of a ball with radius r is: the volume calculation formula of a ball with radius r is:

The formula for calculating the surface area of a ball with radius r is:

Prove:

Certificate:

To prove, to prove.

Make a hemisphere h=r and a cylinder h=r (as shown in figure 1).

∫V column -V cone

= π×r^3- π×r^3/3

=2/3π×r^3

∴ If the conjecture holds, then V column -V cone =V hemisphere.

According to the Zu principle, two three-dimensional figures sandwiched between two parallel planes are cut by any plane parallel to these two planes. If the obtained two cross-sectional areas are equal, then the volumes of the two three-dimensional figures are equal.

If the conjecture holds, two planes: S 1 (circle) =S2 (ring)

1. According to the formula, the plane area cut from the height h of the hemisphere is π× (R2-H2) 0.5 2 = π× (R2-H2).

2. Make a cone with the same base and height as the cylinder: According to the formula, the area of the ring on the right side of the V cone is π× r 2-π× r× h/r = π× (r 2-h 2).

∵π×(r^2-h^2)=π×(r^2-h^2)

∴V column -V cone =V hemisphere

∫V column -V cone = π× r 3-π× r 3/3 = 2/3 π× r 3

∴V hemisphere = 2/3 π× R 3

From the V hemisphere, it can be deduced that V sphere =2×V hemisphere = 4/3× π r 3.

Certificate of completion

Of course, there are many ways to find the volume of a sphere, and the double integral method is easier to understand.

Solution: The integration area is as shown in the figure.

The radius of a circle is r, and there are many ways to find the volume of a sphere. The method of double integration is easier to understand.

Solution: the integration area is shown in the figure, and the radius of the circle is R.