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How to find the volume of a semicircle

Updated February 23, 2018

Because a semicircle is a two-dimensional shape, it has an area rather than volume. You might need to know the area of a semicircle if you want to paint a semicircular area or if you want to lay sod in a semicircular area. To find the area of a semicircle, you need to know the diameter, which is the distance across the semicircle. If you have half a sphere, you can find the volume by finding the area of the whole sphere and dividing by 2.

Measure the distance across the semicircle to find the diameter of the semicircle.

Divide the diameter by 2 to find the radius. For example, if the semicircle diameter equals 14 inches, divide 14 by 2 to get a radius of 7 inches.

Square the radius. In this example, square 7 to get 49.

Multiply the squared radius by 3.14 to get 153.86 square inches.

Divide 153.86 by 2 to find the area of the semicircle. Completing the example, divide 153.86 by 2 to get 76.93 square inches.

Measure the distance across the bottom of the hemisphere to find the diameter.

Divide the diameter by 2 to find the radius. For example, if the hemisphere diameter equals 14 inches, divide 14 by 2 to get a radius of 7 inches.

Cube the radius. Cubing means to multiply the number by itself three times. In this example, multiply 7 times 7 times 7 to get 343 cubic inches.

Multiply the previous result by 4/3. In this example, multiply 343 by 4/3 to get 457.33 cubic inches.

Multiply the previous result by 3.14 to find the volume if you had a full sphere. In this example, multiply 457.33 by 3.14 to get 1,436.03 cubic inches.

Divide the volume of the full sphere by 2 to find the volume of the hemisphere. In this example, divide 1,436.03 by 2 to get 718.02 cubic inches.

Things You'll Need

  • Measuring tape
  • Calculator
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About the Author

Mark Kennan is a writer based in the Kansas City area, specializing in personal finance and business topics. He has been writing since 2009 and has been published by "Quicken," "TurboTax," and "The Motley Fool."