# What is the volume of a basketball?

###### Wiki User

###### February 08, 2018 8:37AM

About 433.526 inches^3

Answer 2

Authors, Matthew McKay Stephens, Tim Jacobs, Mike Dunning

Goal: Calculate the volume of a basketball as accurately as
possible. Objective 1: Calculate the volume of a basketball. Step
One: Find the radius of the basket ball inflated at 8 PSI. To do
this we used a string to measure the circumference of the basket
ball arriving at a measurement of 29 inches. Next we inserted that
number into the formula C = 2πR where C= circumference and R=
Radius. We found that where C= 29inches, R= 4.615 inches. Step Two:
Graph a circle like the one to the right using the radius
calculated in step one. To do this we took the formula X²+Y²=R² and
inserted our value for the radius arriving at the equation Y=±
√21.29-X². Step Three: Set up an equation to calculate the Volume
of the sphere. To do this we used the disc method where V=π
∫(f(x))²-(g(x))² and entered the positive value of our circle as
f(x) and Y=0 as our g(x). We also decided to bound our equation by
0 and 4.615 thereby only using quadrant I of our graph as indicated
in the second picture to the right. This However would only give us
half of the value of the sphere so we multiplied the entire
equation by 2. Bearing all of this in mind we arrived at the
equation V=2π∫ [(√21.29-X²)²-(0)²]. Step Four: Plug and Chug (solve
the equation) V=2π∫[ (√21.29-X²)²-(0)²] V=2π∫ [21.29-X²]
V=2π[(21.29X-(1/3)X³) bounded by 0, 4.615] V=2π (98.25-32.76)
V=2π(65.53) **V=411.74 inch³** Objective 2: Increase accuracy of
measurements by calculating for areas of diminished volume i.e. the
sunken black lines wrapping around the ball. Noting that the
circumference of one black line wrapping around the ball is 28 ½
inches verses the original circumference of the sphere which was 29
inches. Step one: Measurement and observation We used a string and
determined there was 109 inches of ¼ inch thick black space
wrapping around the ball. Step Two: Calculate the Radius of a
circle with a circumference of 28 ½ inches To do this we again used
the formula C = 2πR where C= circumference and R= Radius. We found
that where C= 28 ½ inches, R= 4.536 inches. Step Three: Set up an
equation to calculate the volume of two ¼ inch thick discs, where
one has a radius of 28 ½ inches and the other a radius 29 inches.
To do this we first assumed the function of each Cartesian graph
would be simply Y=X where X is the value of the radius of the disc
being calculated. We then used the disc method where
V=π∫(f(x))²-(g(x))² and entered the radius of the disc being
calculated as our value for f(x) and 0 as our value for g(x).
Lastly we bounded the equations by X= 0, ¼ Step Four: Plug and Chug
(solve the equations) Disc (1) 28 ½ inch circumference
V=π∫[(4.536)²-(0)²] V=π∫[20.575] V= π[(20.5750(X), bound by 0, ¼ ]
V= π(5.14) **V= 16.160 inches³** Disc (2) 29 inch circumference
V=π∫[(4.615)²-(0)²] V=π∫[21.303] V= π[(21.303(X), bound by 0, ¼ ]
V= π(5.623) **V= 16.713 inches³** Step Five: Take the difference
of the two volumes 16.160-16.731**= -.571 inches³** Step Six:
Divide the difference by 28 ½ inches to determine the amount of
volume lost per inch of black line. -.571/(28 ½ )=
**-.0200408208** Step Seven: Multiply amount per inch by total
inches of black line. -.0200408208*109= **-2.184 inches³** Step
Eight: Add that value to the previous total volume to determine a
more accurate volume that has been adjusted for areas of diminished
volume. 411.74-2.184= **409.556 inches³**

it's a 678686868

**Improved Answer:-**

Volume of a ball or sphere = 4/3*pi*radius3