Calculating Cross-Sectional Area

You may encounter situations in which you have a three-dimensional solid shape and need to figure out the area of an imaginary plane inserted through the shape and having borders defined by the boundaries of the solid.

For example, if you had a cylindrical pipe running under your home measuring 20 meters (m) in length and 0.15 m across, you might want to know the cross-sectional area of the pipe.

Cross sections can be perpendicular to the orientation of the axes of the solid if any exist. In the case of a sphere, any cutting plane through the sphere regardless of orientation will result in a disk of some size.

The area of the cross-section depends on the shape of the solid determining the cross-section’s boundaries and the angle between the solid’s axis of symmetry (if any) and the plane that creates the cross section.

Cross-Sectional Area of a Rectangular Solid

The volume of any rectangular solid, including a cube, is the area of its base (length times width) multiplied by its height: V = l × w × h.

Therefore, if a cross section is parallel to the top or bottom of the solid, the area of the cross-section is l × w. If the cutting plane is parallel to one of the two sets the sides, the cross-sectional area is instead given by l × h or w × h.

If the cross-section is not perpendicular to any axis of symmetry, the shape created may be a triangle (if placed through a corner of the solid) or even a hexagon.

Example: Calculate the cross-sectional area of a plane perpendicular to the base of a cube with a volume of 27 m3.

• Since l = w = h for a cube, any one edge of the cube must be 3 m long (since 3

× 3 

× 3 = 27). A cross-section of the type described would therefore be a square 3 m on a side, giving an area of 9 m2.

Cross-Sectional Area of a Sphere

Any theoretical plane placed through a sphere will result in a circle (think about this for a few moments). If you know either the diameter or the circumference of the circle the cross-section forms, you can use the relationships C = 2πr and A = πr2 to obtain a solution.

Example: A plane is rudely inserted through the Earth very close to the North Pole, removing a section of the planet 10 m around. What is the cross-sectional area of this chilly slice of Earth?

• Since C = 2πr = 10 m, r = 10/2π = 1.59 m;  A = πr2= π(1.59)2= 7.96 m2.

References

• Wolfram MathWorld: Cross Section
• MCCKC: Perimeter and Area