Sphere Volume Decrease Calculation

This solution explains how to determine the percentage decrease in a sphere's volume when its radius is reduced significantly.

Sphere Radius-Volume Relationship Explained

The volume ($V$) of a sphere depends on its radius ($r$). The specific formula used in geometry is:

$$V = \frac{4}{3}\pi r^3$$

This formula shows that the volume is proportional to the cube of the radius ($V \propto r^3$). Therefore, any change in the radius affects the volume by the cube of that change factor.

Calculating Sphere Volume Decrease Step-by-Step

1. Original Sphere Radius and Volume: Let's denote the original radius as $r_1$. The original volume ($V_1$) is given by:

   $$V_1 = \frac{4}{3}\pi r_1^3$$

2. Calculating the New Radius: The problem states the radius is decreased by 80 percent. This means the new radius ($r_2$) will be $100\% - 80\% = 20\%$ of the original radius.

   We can express the new radius $r_2$ in terms of $r_1$:

   $$r_2 = r_1 \times (1 - 0.80) = r_1 \times 0.20$$

   So, the new radius is $0.20$ times the original radius.

3. Calculating the New Volume: Using the new radius $r_2$, the new volume ($V_2$) is:

   $$V_2 = \frac{4}{3}\pi r_2^3$$

   Substitute $r_2 = 0.20 r_1$ into the volume formula:

   $$V_2 = \frac{4}{3}\pi (0.20 r_1)^3$$

   Simplify the cubed term:

   $$(0.20 r_1)^3 = (0.20)^3 \times r_1^3 = 0.008 \times r_1^3$$

   Now substitute this back into the volume formula for $V_2$:

   $$V_2 = \frac{4}{3}\pi (0.008 r_1^3)$$

   Rearrange the terms:

   $$V_2 = 0.008 \times \left(\frac{4}{3}\pi r_1^3\right)$$

   Recognize that $\frac{4}{3}\pi r_1^3$ is the original volume $V_1$. So:

   $$V_2 = 0.008 \times V_1$$

   This tells us the new volume is only 8 thousandths (or 0.8%) of the original volume.

4. Calculating the Percentage Decrease: To find the percentage decrease, we use the formula:

   $$\text{Percentage Decrease} = \frac{\text{Original Volume} - \text{New Volume}}{\text{Original Volume}} \times 100\%$$

   Plugging in $V_1$ and $V_2 = 0.008 V_1$:

   $$\text{Percentage Decrease} = \frac{V_1 - 0.008 V_1}{V_1} \times 100\%$$

   Factor out $V_1$ in the numerator:

   $$\text{Percentage Decrease} = \frac{(1 - 0.008) V_1}{V_1} \times 100\%$$

   Cancel out $V_1$:

   $$\text{Percentage Decrease} = (1 - 0.008) \times 100\%$$

   $$\text{Percentage Decrease} = 0.992 \times 100\%$$

   $$\text{Percentage Decrease} = 99.2\%$$

Sphere Volume Decrease Conclusion

When the radius of a sphere is decreased by 80%, the volume decreases by 99.2%. This significant decrease happens because the volume depends on the cube of the radius.