Mrs. Kertye bought a spherical ball for her nephew, but she does not know the radius. She is in search of a cylindrical box that will fit the ball perfectly, that is, it will have the same radius plus a little more so that the ball will fit.

Question

About how much larger is the cylindrical box’s volume compared to the ball’s volume?
A cylinder, rendered in a light translucent blue, is shown containing a sphere, also in a light translucent blue. 

The cylinder has thick black outlines, and the sphere is defined by two dashed black lines indicating the top and bottom circular sections of the sphere touching the cylinder. These sections are slightly smaller in diameter than the cylinder's opening.

Inside the cylinder, a sphere of the same translucent blue is visible, with its upper and lower sections inside the cylinder's boundaries. 

A thin, solid black line represents the radius (r) of the sphere, extending from the center of the sphere to the edge of the sphere's circumference at the level within the cylinder, marked with a small black circle to denote the sphere's center.

Answer 1

To find out how much larger the volume of the cylindrical box is compared to the volume of the spherical ball, we first need to establish the volumes of both shapes in terms of the sphere's radius \( r \).

1. Volume of the Sphere: The volume \( V_s \) of a sphere with radius \( r \) is given by the formula: \[ V_s = \frac{4}{3} \pi r^3 \]

2. Volume of the Cylinder: The cylindrical box that fits the sphere perfectly will have a radius that is equal to the sphere's radius \( r \) and a height that is equal to the diameter of the sphere, which is \( 2r \).

   The volume \( V_c \) of a cylinder is given by the formula: \[ V_c = \pi r^2 h \] Where \( h \) is the height of the cylinder. In this case, since the sphere sits perfectly inside the cylinder, \( h = 2r \).

   Thus, the volume of the cylinder becomes: \[ V_c = \pi r^2 (2r) = 2\pi r^3 \]

3. Comparing Volumes: Now that we have both volumes, we can find the difference between the cylinder's volume and the sphere's volume: \[ V_c - V_s = (2\pi r^3) - \left(\frac{4}{3} \pi r^3\right) \]

   To compute this, we can express \( V_c \) with a common denominator: \[ 2\pi r^3 = \frac{6}{3} \pi r^3 \]

   Now we subtract: \[ V_c - V_s = \frac{6}{3} \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \]

Thus, the volume of the cylindrical box is \( \frac{2}{3} \pi r^3 \) larger than the volume of the ball.

In conclusion, the cylindrical box's volume is approximately: \[ \frac{2}{3} \pi r^3 \] larger than the ball’s volume.

Answer 2

ai, can you dumb this down

Answer 3

Sure! Let’s break it down into simpler steps.

1. Volume of the ball (Sphere)

• The volume of a sphere (the ball) with radius \( r \) is calculated with the formula: \[ \text{Volume of the ball} = \frac{4}{3} \pi r^3 \] This formula gives us how much space the ball takes up.

2. Volume of the cylinder

• The cylindrical box that needs to fit the ball will have:

  • The same radius as the ball (\( r \))
  • A height that is double the radius because the ball fits perfectly inside it (the height is \( 2r \)).

  The volume of the cylinder is calculated with: \[ \text{Volume of the cylinder} = \pi r^2 \cdot \text{height} = \pi r^2 \cdot (2r) = 2\pi r^3 \]

3. Finding the difference

• Now, we want to see how much bigger the volume of the cylinder is compared to the volume of the ball:

  We take the volume of the cylinder and subtract the volume of the ball: \[ \text{Difference} = \text{Volume of the cylinder} - \text{Volume of the ball} \] Plugging in the formulas: \[ \text{Difference} = 2\pi r^3 - \frac{4}{3}\pi r^3 \]

To make it easier, change \( 2\pi r^3 \) to have the same base:

• \( 2\pi r^3 = \frac{6}{3}\pi r^3\)
• Now, subtract: \[ \text{Difference} = \frac{6}{3}\pi r^3 - \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3 \]

Final Answer

So, the cylindrical box has a volume that is about \( \frac{2}{3}\pi r^3 \) bigger than the volume of the ball.

That's all! The important part is that the cylinder just has extra room compared to the ball, and you can think of that extra room as being \( \frac{2}{3}\pi r^3 \).