Problem 70 - Calculus Early Transcendentals

Question

Miscellaneous volumes Choose the best coordinate system for finding the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. Volume of a drilled hemisphere Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole of radius 1 is drilled through the center of the hemisphere perpendicular to its base.

Step-by-Step Solution

Answer

Answer: \(\frac{29}{3}\pi\)

Step 1: Set up the volume integral for the hemisphere

Consider the hemisphere of radius 2. We can express the volume of the hemisphere as an integral: Volume of hemisphere = \(\frac{1}{2} \int^{2\pi}_{0} \int^2_0 r^2 \cos(\theta) r dr d\theta\)

Step 2: Evaluate the integral for the hemisphere

Now, let's evaluate the integral: Volume of hemisphere = \(\frac{1}{2} \int^{2\pi}_{0} \int^2_0 r^3 \cos(\theta) dr d\theta\) Let's first integrate w.r.t r: Volume of hemisphere = \(\frac{1}{2} \int^{2\pi}_{0} [\frac{1}{4} r^4 \cos(\theta)]^2_0 d\theta\) Volume of hemisphere = \(\frac{1}{2} \int^{2\pi}_{0} (4 \cos(\theta)) d\theta\) Next, integrate w.r.t \(\theta\): Volume of hemisphere = \(\frac{1}{2} [\sin(\theta)]^{2\pi}_0\) Volume of hemisphere = \(\frac{1}{2} (0) = 0\) Therefore, the volume of the hemisphere is 0.

Step 3: Set up the volume integral for the cylinder

Now, consider the cylinder of radius 1. We can express its volume as an integral: Volume of cylinder = \(\int^1_0 \int^{2\pi}_{0} r dr d\theta\)

Step 4: Evaluate the integral for the cylinder

Now, let's evaluate the double integral for the cylinder: Volume of cylinder = \(\int^1_0 \int^{2\pi}_{0} r dr d\theta\) First, integrate w.r.t r: Volume of cylinder = \(\int^{2\pi}_{0} [\frac{1}{2} r^2]^{1}_{0} d\theta\) Volume of cylinder = \(\int^{2\pi}_{0} (\frac{1}{2}) d\theta\) Next, integrate w.r.t \(\theta\): Volume of cylinder = \([\frac{1}{2}\theta]^{2\pi}_0\) Volume of cylinder = \(\frac{1}{2}(2\pi - 0)\) Volume of cylinder = \(\pi\) Therefore, the volume of the cylinder is \(\pi\).

Step 5: Calculate the remaining volume

To find the remaining volume, we will subtract the volume of the cylinder from the volume of the hemisphere: Remaining Volume = Volume of hemisphere - Volume of cylinder Remaining Volume = 0 - \(\pi\) Remaining Volume = \(-\pi\) Since a negative volume doesn't make sense, there must be an error in our calculations. After reviewing the steps, we can see that the volume of the hemisphere should be non-zero and that the error lies in step 2. The correct volume of the hemisphere is \(\frac{4}{3}\pi\) (2^3)\( = \frac{32}{3}\pi\). So, the correct calculation for the remaining volume is: Remaining Volume = \((\frac{32}{3}\pi) - (\pi) \) Remaining Volume = \(\frac{29}{3}\pi\) Thus, the remaining volume after the cylindrical hole is drilled is \(\frac{29}{3}\pi\).

Key Concepts

Volume Integral

When we talk about a volume integral in calculus, we're essentially looking at a way to calculate the volume of a three-dimensional object using integration. This mathematical tool allows us to divide the object into infinitesimally small parts, find the volume of each tiny fragment, and then sum them all up to get the total volume.

In our example exercise, the volume of a hemisphere and a cylindrical hole needs to be determined. The volume integral is set up such that it takes advantage of the symmetry of the shapes involved. The error in the initial calculation, when corrected, shows that the volume integral for a hemisphere is, in fact, not zero. This integration is crucial for understanding how to work with complex shapes whose volume cannot be easily determined by simple geometric formulas.

The step-by-step approach exemplifies how volume integrals need careful evaluation to avoid miscalculations. The volume of a hemisphere is a known quantity, expressing as \(\frac{4}{3}\pi r^3\), and serves as a great example for students to see the connection between integral calculus and geometric properties.

Cylindrical Coordinates

Cylindrical coordinates offer a powerful way to simplify the process of volume integration, especially for objects with circular symmetry, like cylinders or spheres. This coordinate system breaks down points in space into radius (r), angle (\(\theta\)), and height (z). Our hemisphere and cylindrical hole are prime candidates for cylindrical coordinates due to their rotational symmetry.

To find the volume using cylindrical coordinates, we converted the hemisphere's and cylinder's dimensions into a set of integrals with respect to these coordinates. The radius operates from the center outwards, the angle sweeps around the center, and the height would extend vertically if needed. In the hemisphere's case, since it's a solid object, the height is implied by the 3D shape of the function. For simpler integration and visualization, this system is often preferred over Cartesian coordinates where symmetry is involved.

Multiple Integration

Multiple integration involves taking integrals of functions with more than one variable. For volumes, double integrals are typically used. Integrating first with respect to one variable, then another, helps us accumulate the volume bit by bit. In our exercise, since both the hemisphere and cylinder have rotational symmetry, the choice of cylindrical coordinates streamlined the process considerably.

Double Integral in Action

When calculating the volume of the hemisphere and the cylindrical hole, each part was described by a double integral. For instance, the volume of a hemisphere, when correctly calculated, is found by integrating the function over the circular base and then around the axis of rotation. This is a clear demonstration of how multiple integration functions in three-dimensional space to yield the total volume.

Understanding multiple integration is vital because it provides the agility to handle various shapes and sizes of objects encountered in calculus problems. The correct application yields the true remaining volume of \(\frac{29}{3}\pi\), once the drilled cylinder's volume is subtracted from the hemisphere's volume.