Matthew Stephens and Group Complete Sports and STEM Project

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 basketball 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 28½ inches vs. 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   The Volume of a Basketball These calculations include the outer skin of the ball as well as any air trapped inside the ball.             Matthew McKay Stephens Michael Dunning Tim Jacobs