A baseball has a circumference of 9 inches. A softball has a circumference of 12 inches. What is the approximate difference in volume between the two balls? Round your answer to the nearest hundredths place.
+26367 A baseball has a circumference of 9 inches.
A softball has a circumference of 12 inches.
What is the approximate difference in volume between the two balls?
Round your answer to the nearest hundredths place.
\(\text{Let $r_b = $ radius baseball }\\ \text{Let $r_s = $ radius softball }\)
1. baseball
\(9 = 2\pi r_b \\ r_b = \dfrac{9}{2\pi} \\ V_b = \dfrac{4}{3}\pi r_b^3 \)
2. softball
\(12 = 2\pi r_s \\ r_s = \dfrac{12}{2\pi} \\ V_s = \dfrac{4}{3}\pi r_s^3 \)
3. difference in volume
\(\begin{array}{|rcll|} \hline V_s - V_b &=& \dfrac{4}{3}\pi r_s^3 - \dfrac{4}{3}\pi r_b^3 \\\\ &=& \dfrac{4}{3}\pi\left( r_s^3 - r_b^3 \right) \\\\ &=& \dfrac{4}{3}\pi\left[ \left(\dfrac{12}{2\pi} \right)^3 - \left(\dfrac{9}{2\pi} \right)^3 \right] \\\\ &=& \dfrac{4\pi}{3\cdot 8\pi^3} \left( 12^3 - 9^3 \right) \\\\ &=& \dfrac{1}{3\cdot 2\pi^2} \cdot 999 \\\\ &=& \dfrac{166.5}{\pi^2} \\\\ &=& 16.8699770764 \\ \hline \end{array}\)
\(\text{The approximate difference in volume between the two balls is $\mathbf{16.87\ in^3}$ }\)
