1. The figure shows a solid which consists of a hemisphere and a right circular cylinder with the same base. The height of the circular cylinder is twice its base radius. If the volume of the solid is \( \frac{512 \pi}{3} \mathrm{~cm}^{3} \), find the height of the circular cylinder. 2. In the figure, the radius of the sphere and the base radius of the right circular cylinder are the same. The base radius and the height of the circular cylinder are the same. If the total volume of the two solids is \( 63 \pi \mathrm{~cm}^{3} \), find the diameter of the sphere.
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To solve the first problem, we recognize that the volume of the solid consists of the volume of the hemisphere and the cylinder. The volume \( V \) of a hemisphere is given by \( \frac{2}{3} \pi r^3 \), and the volume of a cylinder is \( \pi r^2 h \). Given that the height \( h \) of the cylinder is twice the base radius \( r \), we can substitute \( h = 2r \) into the volume formula. So, the total volume becomes: \[ V = \frac{2}{3} \pi r^3 + \pi r^2 (2r) = \frac{2}{3} \pi r^3 + 2\pi r^3 = \frac{2}{3} \pi r^3 + \frac{6}{3} \pi r^3 = \frac{8}{3} \pi r^3 \] We set this equal to the given volume \( \frac{512 \pi}{3} \): \[ \frac{8}{3} \pi r^3 = \frac{512 \pi}{3} \] Solving for \( r^3 \), we find \( 8r^3 = 512 \), which gives \( r^3 = 64 \), leading to \( r = 4 \) cm. The height of the cylinder is: \[ h = 2r = 2 \times 4 = 8 \text{ cm.} \] For the second problem, we have a sphere and a cylinder with the same radius \( r \) and the cylinder's height \( h = r \). The volume of the sphere is \( \frac{4}{3} \pi r^3 \), and the volume of the cylinder is \( \pi r^2 h = \pi r^3 \). Thus, the total volume is: \[ V = \frac{4}{3} \pi r^3 + \pi r^3 = \frac{4}{3} \pi r^3 + \frac{3}{3} \pi r^3 = \frac{7}{3} \pi r^3 \] Setting this equal to \( 63 \pi \): \[ \frac{7}{3} \pi r^3 = 63 \pi \] Dividing both sides by \( \pi \) and multiplying by \( \frac{3}{7} \): \[ r^3 = 63 \times \frac{3}{7} = 27 \] Thus, \( r = 3 \) cm. The diameter of the sphere is \( 2r = 2 \times 3 = 6 \) cm.