Prisms and Cylinders
Volume = base area × height
Cylinder: volume = πr²h、lateral area = 2πrh、surface area = 2πr²+2πrh
Pyramids and Cones
Volume = (1/3) × base area × height
Cone: volume = (1/3)πr²h、slant height l = √(r²+h²)、lateral area = πrl
Sphere
Volume = (4/3)πr³、surface area = 4πr²
Example ① (Cone and sphere)
For a cone with base radius 3 and height 4:
Slant height l = √(9+16) = 5
Volume = (1/3)π×9×4 = 12π
Lateral area = π×3×5 = 15π
For a cone with base radius 3 and height 4:
Slant height l = √(9+16) = 5
Volume = (1/3)π×9×4 = 12π
Lateral area = π×3×5 = 15π
Example ② (Cylinder inscribed in a sphere)
For a cylinder inscribed in a sphere of radius R, let its height be h and base radius be r. Then r² + (h/2)² = R²
When R=3 and h=4, r² = 9 − 4 = 5 → r = √5
Volume = π×5×4 = 20π
For a cylinder inscribed in a sphere of radius R, let its height be h and base radius be r. Then r² + (h/2)² = R²
When R=3 and h=4, r² = 9 − 4 = 5 → r = √5
Volume = π×5×4 = 20π
Example ③ (Cube circumscribed about a sphere)
When the radius of a sphere is r, the edge length of a cube circumscribed about the sphere is 2r.
The radius of a sphere inscribed in a cube is r = a/2 when the edge length is a.
When the radius of a sphere is r, the edge length of a cube circumscribed about the sphere is 2r.
The radius of a sphere inscribed in a cube is r = a/2 when the edge length is a.
Key points
- The volume of a pyramid or cone is 1/3 that of a prism or cylinder
- Sphere volume (4/3)πr³、surface area 4πr²
- Use the Pythagorean theorem for inscribed and circumscribed relationships between a sphere and other solids
Practice Problems
- Find the volume and surface area of a sphere with radius 6.
- Find the volume and slant height of a cone with base diameter 8 and height 6.
- Find the volume of a sphere inscribed in a cube with edge length 6.
Answers and Explanations
- Volume=(4/3)π×216=288π、surface area=4π×36=144π
- r=4、h=6。Volume=(1/3)π×16×6=32π。Slant height=√(16+36)=√52=2√13
- Radius of the inscribed sphere=3（half the cube's edge）. Volume=(4/3)π×27=36π