FINDING THE

VOLUME OF IRREGULARLY

SHAPED SOLIDSby Moira Whitehouse PhD

¥ SOME REGULAR SHAPES IN WHICH SOLIDS OCCUR INCLUDE:

sphere

rectangular prisms

cube

cone

cylinder

pyramid

¥A MATHEMATICAL FORMULA CAN BE USED TO FIND THE VOLUME OF ANY OF THESE SOLIDS AS LONG THE OBJECTÕS LENGTH, WIDTH, AND HEIGHT, AND FOR SOME, DIAMETER, BASE OR RADIUS ARE KNOWN.

¥FOR EXAMPLE, THE VOLUME FORMULA FOR A RECTANGULAR PRISM IS ITS LENGTH x ITS WIDTH x ITS HEIGHT.

LWH

¥AND THE FORMULA FOR THE VOLUME OF A CYLINDER SAYS MULTIPLY 3.14 x ITS RADIUS x ITS RADIUS x ITS HEIGHT

HEIGH

¥ALTHOUGH USING FORMULAS TO FIND VOLUMES OF SOLID OBJECTS IS FUN, YOU WONÕT BE DOING MUCH OF THAT UNTIL YOU ARE IN 7TH OR 8TH GRADE.

¥HOWEVER, YOU WILL FIND THAT MOST SOLIDS DO NOT HAVE A REGULAR SHAPE--RATHER THEY HAVE IRREGULAR SHAPES.

¥HOWEVER, YOU WILL FIND THAT MOST SOLIDS DO NOT HAVE A REGULAR SHAPE--RATHER THEY HAVE IRREGULAR SHAPES.

¥EVEN THE ROUNDEST ROCK IS NOT A PERFECT SPHERE.

¥IN GENERAL SOLIDS SUCH AS MINERALS, PLANTS, ANIMALS, FEATHERS, STOMACHS AND SO FORTH HAVE THE MOST PECULIAR SHAPES AND OFTEN EVEN HAVE PARTS STICKING OUT FROM THEM.

¥SO HOW CAN SCIENTISTS POSSIBLY FIND THE VOLUME OF OBJECTS LIKE THESE?

¥THEY MEASURE THE VOLUME OF SUCH SOLIDS BY A METHOD CALLED WATER DISPLACEMENT.

¥SO HOW CAN SCIENTISTS POSSIBLY FIND THE VOLUME OF OBJECTS LIKE THESE?

¥THEY PLACE THE SOLID IN WATER AND MEASURE HOW MUCH WATER IS PUSHED OUT OF THE WAY (DISPLACED).

¥IF AN OBJECT IS PLACED IN A CONTAINER THAT IS FULL OF WATER, THE WATER WILL RUN OVER THE TOP.

¥THEY PLACE THE SOLID IN WATER AND MEASURE HOW MUCH WATER IS PUSHED OUT OF THE WAY (DISPLACED).

FULL ADD ICEOVERFLOW!

¥OR PLACE AN OBJECT IN A CONTAINER OF WATER THAT IS NOT FULL, AND THE WATER LEVEL WILL RISE.

ADDING ICE

¥THIS IS BECAUSE THE ICE TAKES UP THE SPACE THAT WATER WAS IN AND PUSHES THE WATER OUT OF THE WAY.

¥THE AMOUNT OF WATER THAT IS PUSHED OUT OF THE WAY HAS THE SAME VOLUME AS THE ICE.

¥THIS IS BECAUSE THE ICE TAKES UP THE SPACE THAT WATER WAS IN AND PUSHES THE WATER OUT OF THE WAY.

¥THIS INFORMATION CAN BE USED TO HELP FIND THE VOLUME OF SOLID OBJECTS.

¥WE WILL USE A CONTAINER WITH A SPOUT.

¥THIS INFORMATION CAN BE USED TO HELP FIND THE VOLUME OF SOLID OBJECTS.

¥FILL THE CONTAINER TO THE LEVEL OF THE SPOUT.

¥PUT THE OBJECT IN THE CONTAINER,

¥FILL THE CONTAINER TO THE LEVEL OF THE SPOUT.

¥THE WATER THAT FLOWS OUT EQUALS THE VOLUME OF THE

OBJECT.¥THUS, BY MEASURING THE WATERÕS VOLUME, WE WILL KNOW THE OBJECTÕS VOLUME.

¥THE DISPLACED WATERÕS VOLUME IS EASILY READ INA GRADUATED CYLINDER.

¥THUS, BY MEASURING THE WATERÕS VOLUME, WE WILL KNOW THE OBJECTS VOLUME.

¥IF THE AMOUNT OF WATER THAT FLOWED OUT OF THE SPOUT WAS 100 mL THE ROCKÕS VOLUME IS 100 mL OR 100 cc (BECAUSE ONE CUBIC CENTIMETER [cc] AND ONE MILLILITER [mL] TAKE UP THE SAME AMOUNT OF SPACE)

¥WE COULD ALSO MEASURE HOW MUCH THE WATER RISES WHEN THE OBJECT IS PLACED IN A CONTAINER.

¥DISPLACEMENT CAN BE USED IN THIS MANNER TO FIND THE VOLUME OF SMALLER OBJECTS.

¥WE COULD ALSO MEASURE HOW MUCH THE WATER RISES WHEN THE OBJECT IS PLACED IN A CONTAINER.

¥SUPPOSE WE WANT TO MEASURE THE VOLUME OF THIS ROCK.

¥BEFORE PLACING THE ROCK IN THE BEAKER, NOTICE THAT THE WATER LEVEL IS500 mL.

¥WHEN WE PLACE THE ROCK IN THE BEAKER, THE WATER LEVEL RISES TO 600 mL.

¥WE CAN SUBTRACT THESE TWO NUMBERS TO GET THE VOLUME OF THE ROCK.

¥600 mL (VOLUME OF ROCK AND WATER) MINUS 500 mL (VOLUME OF WATER) EQUALS THE VOLUME OF THE ROCK (100 mL)

¥WE CAN SUBTRACT THESE TWO NUMBERS TO GET THE VOLUME OF THE ROCK.

¥SINCE THE VOLUME OF A CUBIC CENTIMETER AND A MILLILITER ARE THE SAME, WE CAN AGAIN SAY THAT THE ROCK HAS A VOLUME OF 100 cc.

¥BEFORE WE PLACE THE RECTANGULAR PRISM IN A BEAKER OF WATER TO MEASURE ITS VOLUME, WE MUST NOTE THE WATERLEVEL--600 mL.

¥WHEN WE PLACE IT IN THE BEAKER OF WATER, WE SEE THAT THE WATER LEVEL RISES TO700 mL SO WHAT IS THE PRISMS VOLUME?

THIS ROCK AND THIS RECTANGULAR PRISM, THEREFORE HAVE THE SAME VOLUME -- 100 cc or 100 mL.

TO MEASURE THE VOLUME OF IRREGULARLY SHAPED OBJECTS, SCIENTISTS USE THE PROCEDURE CALLED DISPLACEMENT. THE VOLUME OF THE OBJECT EQUALS THE AMOUNT OF WATER IT DISPLACES.