Answer: A
If the radius of the sphere is \( r \) units, then:
The total surface area \( A \) is given by \( A = 4\pi r^2 \), which leads to \( r^2 = \frac{A}{4\pi} \).
The volume \( V \) of the sphere is given by \( V = \frac{4}{3} \pi r^3 \), which leads to \( r^3 = \frac{3V}{4\pi} \).
By cubing both equations:
\( r^6 = \frac{A^3}{64\pi^3} \)
\( r^6 = \frac{9V^2}{16\pi^2} \)
Equating the two expressions for \( r^6 \): \[ \frac{A^3}{64\pi^3} = \frac{9V^2}{16\pi^2} \] Simplifying this, we get: \[ \frac{A^3}{4\pi} = 9V^2 \] Finally: \[ A^3 = 36\pi V^2 \] So, the relationship between the total surface area \( A \) and the volume \( V \) of the sphere is \( A^3 = 36\pi V^2 \).
If the radius of the sphere is \( r \) units, then:
The total surface area \( A \) is given by \( A = 4\pi r^2 \), which leads to \( r^2 = \frac{A}{4\pi} \).
The volume \( V \) of the sphere is given by \( V = \frac{4}{3} \pi r^3 \), which leads to \( r^3 = \frac{3V}{4\pi} \).
By cubing both equations:
\( r^6 = \frac{A^3}{64\pi^3} \)
\( r^6 = \frac{9V^2}{16\pi^2} \)
Equating the two expressions for \( r^6 \): \[ \frac{A^3}{64\pi^3} = \frac{9V^2}{16\pi^2} \] Simplifying this, we get: \[ \frac{A^3}{4\pi} = 9V^2 \] Finally: \[ A^3 = 36\pi V^2 \] So, the relationship between the total surface area \( A \) and the volume \( V \) of the sphere is \( A^3 = 36\pi V^2 \).