Problem 51 - Thomas Calculus

Question

Volume of a bowl a. A hemispherical bowl of radius \(a\) contains water to a depth \(h .\) Find the volume of water in the bowl. b. Related rates Water runs into a sunken concrete hemi- spherical bowl of radius 5 \(\mathrm{m}\) at the rate of \(0.2 \mathrm{m}^{3} / \mathrm{sec} .\) How fast is the water level in the bowl rising when the water is 4 \(\mathrm{m}\) deep?

Step-by-Step Solution

Answer

a. Volume: \( \pi(a^2h - \frac{h^3}{3}) \) cubic meters. b. Rate: \( \frac{dh}{dt} \approx 0.0071 \; \text{m/s} \) at 4 m depth.

Step 1: Understanding the Volume of a Hemisphere

The volume, \( V \), of a full hemisphere of radius \( a \) is given by the formula:\[ V = \frac{2}{3} \pi a^3 \] We need to find the volume of a hemispherical bowl partially filled to a depth \( h \).

Step 2: Calculating Volume of Water at Depth h

The volume of water in a hemisphere to a depth \( h \) can be found by integrating horizontal slices from the bottom to depth \( h \). The radius of each slice at depth \( x \) is \( \sqrt{a^2 - x^2} \). Thus, the volume is:\[ V = \pi \int_0^h (a^2 - x^2) \, dx = \pi \left[ a^2x - \frac{x^3}{3} \right]_0^h = \pi \left( a^2h - \frac{h^3}{3} \right) \]

Step 3: Understanding Related Rates Problem

We need to determine how fast the water level is rising in the hemisphere when the water depth is 4 m. We know the rate of change of volume, \( \frac{dV}{dt} = 0.2 \; \text{m}^3/\text{s} \) and need to find \( \frac{dh}{dt} \) when \( h = 4 \; \text{m} \).

Step 4: Differentiate Volume Formula with Respect to Time

Start from the volume formula derived in Step 2:\[ V(h) = \pi \left( a^2h - \frac{h^3}{3} \right) \]Differentiate both sides with respect to \( t \):\[ \frac{dV}{dt} = \pi \left( a^2 \frac{dh}{dt} - h^2 \frac{dh}{dt} \right) \]So,\[ \frac{dV}{dt} = \pi (a^2 - h^2) \frac{dh}{dt} \]

Step 5: Substituting Known Values to Solve for dh/dt

Substitute \( \frac{dV}{dt} = 0.2 \), \( a = 5 \), and \( h = 4 \) into the differentiated formula:\[ 0.2 = \pi (5^2 - 4^2) \frac{dh}{dt} \]\[ 0.2 = \pi (25 - 16) \frac{dh}{dt} \]\[ 0.2 = 9\pi \frac{dh}{dt} \]Solving for \( \frac{dh}{dt} \), we get:\[ \frac{dh}{dt} = \frac{0.2}{9\pi} \approx 0.0071 \; \text{m/s} \]

Step 6: Conclusion

The volume of water in the bowl when filled to a depth \( h \) is \( \pi(a^2h - \frac{h^3}{3}) \) cubic meters, and the water level rises at a rate of approximately 0.0071 m/s when the water depth is 4 m.

Key Concepts

Volume of a Hemisphere

To find the volume of a hemisphere, we begin with its geometric nature. A hemisphere is essentially half of a sphere. If you picture a full sphere, and split it through its center, each half is a hemisphere. This means calculating the volume of a hemisphere involves dividing the sphere's formula by two.
The volume of a full sphere is given by the equation \( V = \frac{4}{3} \pi r^3 \). For a hemisphere, this becomes \( V = \frac{2}{3} \pi r^3 \). In this problem, we are dealing with a hemisphere partially filled with water, so we must adjust our calculations to find the volume up to a certain depth, \( h \).

In real-world scenarios, such as finding the volume of water in a bowl or tank, we often use integration to calculate the volume precisely. This ensures we consider the curves and shape of the object.

Related Rates

Related rates problems involve finding the rate at which one quantity changes in relation to another. These are real-life applications of differentiation and very applicable in understanding real-world dynamic systems.

In our exercise, the water is filling a hemispherical bowl, and we are tasked with determining how fast the water level rises over time. We know how fast water is being added, namely \( \frac{dV}{dt} = 0.2 \ m^3/s\), and we need to find how its depth, \( h \), changes relative to time, \( \frac{dh}{dt} \).
This is a classic example of a related rates problem. We use differentiation to relate how volumes and heights change with respect to time. Recognizing and setting up these relationships help in understanding how different measurements influence one another in a dynamic scenario.

Integration

Integration is a fundamental concept in calculus used for finding areas, volumes, and sums. It is applied in this exercise to find the volume of water in the hemisphere up to a certain depth.

To determine the volume of water in a hemisphere to depth \( h \), we don’t just use the volume formula for a whole hemisphere but instead integrate horizontal slices from the bottom to the depth \( h \).
Each slice at depth \( x \) has a radius \( \sqrt{a^2 - x^2} \). The volume of each of these infinitesimally thin slices gives us the integral: \[ V = \pi \int_0^h (a^2 - x^2) \, dx \]
Switching from sums of a series of slices to one continuous integral is one of the powerful techniques that make calculus so effective. This particular integral helps us find the precise volume of a variable-shaped region, like a hemisphere.

Differentiation

Differentiation is a key operation in calculus, allowing us to determine how a function changes as its inputs change. In related rates problems, differentiation helps relate time-dependent changes in variables.

From our volume expression derived through integration, \( V(h) = \pi (a^2h - \frac{h^3}{3}) \), we compute its derivative with respect to time to understand how changes in water level impact its volume. This yields the formula: \(\frac{dV}{dt} = \pi (a^2 - h^2) \frac{dh}{dt} \)

This formula connects water volume changes to height changes, letting us find \( \frac{dh}{dt} \) given \( \frac{dV}{dt} \). Solving correctly by substituting the known values allows us to understand how quickly the water level in the hemisphere rises as water continues to flow in.