Problem 67 - Calculus Early Transcendentals

Question

A hemispherical bowl of radius 8 inches is filled to a depth of \(h\) inches, where \(0 \leq h \leq 8\) ( \(h=0\) corresponds to an empty bowl). Use the shell method to find the volume of water in the bowl as a function of \(h\). (Check the special cases \(h=0\) and \(h=8 .)\)

Step-by-Step Solution

Answer

Question: Using the shell method, find the volume of water in a hemispherical bowl with radius 8 inches when it is filled to a depth of h inches. Answer: The volume of water in the bowl when filled to a depth of h inches is given by the integral: \(-\pi \left[\frac{2}{3} u^{\frac{3}{2}}\right]_{64}^0\), which evaluates to \(\pi \frac{256}{3}\) cubic inches for the hemispherical bowl of radius 8 inches.

Step 1: Visualize the Problem

First, let's visualize the hemispherical bowl and the water volume inside it. The radius of the bowl is given as 8 inches. The hemisphere lies on the x-y plane in a coordinate system, with z representing the height or depth of water in the bowl. In this problem, we will only consider the heights where 0 ≤ h ≤ 8.

Step 2: Set up the Shell Method

The shell method involves integrating over a single parameter, in this case, the radius of the cylindrical shells, \(r\). The volume of a thin cylindrical shell with height h is given by \(2\pi rhd(r)\), and we will consider shells with varying radii from 0 to 8, to encompass the entire volume in the bowl.

Step 3: Calculate the Height of the Shell

We want to find the connection between the height \(h\) and the radius of the cylindrical shells. Since the bowl is a hemisphere, we know that the height varies from 0 to 8 inches as we move from the center to the edges of the hemisphere. We can use the equation of a circle, \(r^2 + h^2=8^2\), to find the relationship between the height and the radius, which gives us \(h=\sqrt{(8^2-r^2)}\). This is the height function we will use when integrating.

Step 4: Integrate to Find the Volume

Now, we can integrate the volume of the cylindrical shells over the range of the radii (0 to 8) to find the volume of water in the bowl with respect to the depth \(h\). \(\int_0^8 2\pi r\sqrt{64-r^2} dr\) By using a substitution, let \(u =64-r^2\), and \(du = -2r dr\). Then, convert the limits of integration. If \(r=0\), then \(u=64\) and if \(r=8\), then \(u=0\). The new integral becomes: \(-\pi \int_{64}^0 u^{\frac{1}{2}} du\)

Step 5: Evaluate and Solve the Integral

Now, we can solve the integral to obtain the volume function of water in the bowl. \(-\pi \left[\frac{2}{3} u^{\frac{3}{2}}\right]_{64}^0\) = \(\pi \left[\frac{2}{3} (64)^{\frac{3}{2}} - \frac{2}{3}(0)^{\frac{3}{2}}\right]\) = \(\pi \frac{256}{3}\) cubic inches

Step 6: Check Special Cases

We need to check the volume of water when h=0 and when h=8 (maximum capacity). When \(h=0\), all shells have a height of 0, and the volume of water should be 0 as well. When \(h=8\), all shells should have their maximum height, which means they have completely filled the hemisphere. Indeed, the volume of water in the bowl when h=8 is the same as the hemisphere's volume: Volume of hemisphere = \(\frac{2}{3}\pi(8)^3=\pi \frac{256}{3}\) cubic inches.

Key Concepts

Hemisphere Volume

Understanding the volume of a hemisphere is crucial for solving problems involving a hemispherical bowl or any half-spherical container. To find the volume of a hemisphere with a radius, say 8 inches, you start by considering the volume of a full sphere and then divide it by two.

For a full sphere, the formula for volume is \ \( V = \frac{4}{3} \pi r^3 \ \). Since a hemisphere is half of that sphere, the volume becomes \ \( V = \frac{2}{3} \pi r^3 \ \). It's helpful to derive this from the fundamental sphere volume to cement the understanding of how much space half of a sphere occupies. This fundamental concept underpins the more complex calculus required to compute variable water levels in hemispherical containers.

Integration Techniques

In calculus, integration techniques allow us to find areas, volumes, and other quantities when only their derivatives are known. The integral we used in this problem was \ \( \int_0^8 2\pi r\sqrt{64-r^2} dr \ \).

There are different methods to solve integrals. Here, we used a substitution strategy to simplify the integral.

• First, identify a substitution that makes the integral easier to solve. We chose \ \( u = 64 - r^2 \ \).
• Differentiating gives \ \( du = -2r \, dr \ \).
• This substitution simplifies the integral into a form that's easier to evaluate.

By practicing these techniques, you sharpen your calculus skills and prepare better for more complicated problems.

Cylindrical Shells

The shell method provides a solution to find volumes of revolution by imagining the volume built from numerous cylindrical shells. In our bowl problem, the idea is to stack shells of different heights and radii to fill the bowl.

• The shell's height relates to the water level, identified by \ \( h = \sqrt{64-r^2} \ \) in our problem.
• Create each shell by revolving strips around an axis outside the curve defining the region.

The integral of the shell method provides the total volume, stacking each infinitesimally thin shell between the given limits. This method is especially useful for volumes where cross-sectional areas vary.”.

Calculus Problem Solving

Solving calculus problems can seem daunting but breaking them into smaller steps simplifies the process. Here's how we approached our problem:

• **Visualize & understand the problem:** We looked at the bowl and started by asking the right questions.
• **Setup the integral:** We created a plan by determining what to integrate and which bounds to use.
• **Perform substitutions:** This was crucial for simplification, turning the problem into more solvable terms.
• **Solve and check special cases:** We verified our results against special cases to ensure correctness, e.g., when water is empty or full.

Always remember, each solution in calculus stands on these foundational problem-solving principles.