Question

Estimating Volume You can estimate the volume of a sphere by measuring its circumference with a tape measure, dividing by 2\(\pi\) to get the radius, then using the radius in the volume formula. Find how sensitive your volume estimate is to a 1\(/ 8\) in. error in the circumference measurement by filling in the table below for spheres of the given sizes. Use differentials when filling in the last column. $$\begin{array}{|c|c|c|}\hline \text { Sphere Type } & {\text { True Radius }} & {\text { Tape Error Radius Error Volume Error }} \\ \hline \text { Orange } & {2 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Melon } & {4 \text { in. }} & {1 / 8 \text { in. }} \\ \hline \text { Beach Ball } & {7 \text { in. }} & {1 / 8 \text { in. }}\end{array}$$

Short answer

The estimated volumes and their errors are obtained using the differential formula. They provide an estimate of the sensitivity of volume calculation to errors in circumference measurement.

Step-by-step solution

Find Estimated Radii

First, calculate the estimated radius by adding the measurement error to the true radius for each sphere: \[ r_{\text{est}} = r_{\text{true}} + \frac{1}{8} \]

Calculate Volume of the Spheres

Then, calculate the volume V of each type of sphere using the volume formula for a sphere with the estimated radii: \[ V = \frac{4}{3} \pi r_{\text{est}}^3 \]

Calculate Volume With True Radii

Also compute the volume using the true radius: \[ V_{\text{true}} = \frac{4}{3} \pi r_{\text{true}}^3 \]

Calculate the Volume Error

Subtract the volume with the true radius from the one estimated with the radius error. This will give the error in the volume estimation: \[ V_{\text{err}} = |V - V_{\text{true}}| \]

Use Differentials for Calculating Volume Error

We can also calculate the volume error using differentials. Differentiate the volume formula with respect to radius, then multiply by the radius error (which is 1/8 in. in this case): \[ \Delta V = \frac{\partial V}{\partial r} \cdot \Delta r = 4 \pi r_{\text{true}}^2 \cdot \frac{1}{8} \]

Key concepts

Spherical Volume Calculation

Understanding how to calculate the volume of a sphere is essential when dealing with three-dimensional shapes in geometry. The volume of a sphere can be found using the formula
\[ V = \frac{4}{3} \pi r^3 \]
where \( V \) represents the volume and \( r \) is the radius of the sphere. This formula comes from the integral calculus and it's a result of revolving a semicircle around its diameter.
When measuring a real-world object to approximate its volume, you would typically measure its circumference and then divide by \( 2\pi \) to get the radius. This radius can then be plugged into the volume formula. An accurate measurement of the circumference is crucial because even a small error can lead to a significant volume estimation error, especially for larger spheres. When working with volume calculations, it's helpful to visualize the sphere and understand the relationship between radius and volume – a small change in radius results in a larger change in volume, as the radius affects the volume cubically.

Using Differentials in Error Estimation

Differentials are a powerful tool in calculus, particularly when estimating the effects of measurement errors in volume calculations. Using differentials allows for an approximation of how a small change in the radius, \( \Delta r \), affects the volume change, \( \Delta V \). This relationship can be expressed as:
\[ \Delta V \approx \frac{\partial V}{\partial r} \cdot \Delta r \]
In this equation, \( \frac{\partial V}{\partial r} \) is the derivative of the volume with respect to the radius and represents how sensitive the volume is to changes in the radius. For a sphere, the derivative is \( 4\pi r^2 \), highlighting the squared relationship between the radius and the volume change.
The error in volume, therefore, can be approximated by multiplying the derivative of volume by the error in the radius measurement. This method provides a good approximation of volume error and is particularly useful when exact calculations are complex or impractical.

Sensitivity to Measurement Errors

Sensitivity to measurement errors speaks to how a small change in a measurement can lead to a significant change in the result. This concept is particularly important in cases where precise measurements are difficult to obtain. The volume of a sphere, being proportional to the cube of the radius, is highly sensitive to measurement errors.
For instance, if you measure the circumference of a sphere inaccurately, even by a small amount, converting this measurement to a radius and then calculating volume amplifies the initial error. This is why precision in measuring and calculating is so crucial in practical applications, such as engineering and manufacturing, where an error in volume estimation could lead to significant material waste or design flaws.
As demonstrated in the exercise, even a slight error in measuring the circumference, in this case, 1/8 inch, can cause a noticeable volume error after the calculation. By understanding the concepts of differential calculus and the relationships between a shape's dimensions and its volume, you can better account for and minimize the effects of these measurement sensitivities.