Problem 29 - Calculus Early Transcendentals

Question

Use cylindrical coordinates to find the volume of the following solids. The solid bounded by the plane \(z=0\) and the hyperboloid \(z=\sqrt{17}-\sqrt{1+x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

To find the volume of the solid bounded by the plane \(z=0\) and the hyperboloid \(z=\sqrt{17}-\sqrt{1+x^{2}+y^{2}}\), we first convert the given function to cylindrical coordinates: \(z = \sqrt{17} - \sqrt{1+r^2}\). Then, we set up a triple integral in cylindrical coordinates to find the volume: \(V = \int_{0}^{\sqrt{16}}\int_{0}^{2\pi}\int_{0}^{\sqrt{17} - \sqrt{1+r^2}} r\, dz\, dr\, d\theta\). After evaluating the integral, we find the volume of the solid to be \(\frac{68\pi}{3}\).

Step 1: Convert to Cylindrical Coordinates

To convert the given function \(z=\sqrt{17}-\sqrt{1+x^{2}+y^{2}}\) into cylindrical coordinates, we will replace \(x^2 + y^2\) with \(r^2\). So the equation becomes: \(z = \sqrt{17} - \sqrt{1+r^2}\)

Step 2: Set up the Triple Integral for Volume

To find the volume of the solid, we will set up a triple integral in cylindrical coordinates: \(\displaystyle V = \int\int\int dV\) In cylindrical coordinates, \(dV = r\, dz\, dr\, d\theta\). So, we get: \(\displaystyle V = \int\int\int r\, dz\, dr\, d\theta\)

Step 3: Determine the Limits of Integration

Next, we need to find the limits of integration for our variables. Let's examine the solid's boundaries: 1. The solid is bounded by the plane \(z = 0\). This sets our lower limit for \(z\): \(z = 0\) 2. The solid is bounded by the hyperboloid \(z = \sqrt{17} - \sqrt{1+r^2}\). This sets our upper limit for \(z\): \(z = \sqrt{17} - \sqrt{1+r^2}\) 3. In cylindrical coordinates, \(r\) represents the radius of the circle formed by the projection of the solid onto the xy plane. The projection of our solid onto the xy plane is a circle with radius \(\sqrt{16}\), which means the limits for \(r\) are: \(0 \leq r \leq \sqrt{16}\) 4. The variable \(\theta\) represents the angle of the circle formed by the projection of the solid onto the xy plane. As the solid is symmetric around the z-axis, limits for \(\theta\) will be: \(0 \leq \theta \leq 2\pi\) Now we can rewrite our integral with these limits: \(\displaystyle V = \int_{0}^{\sqrt{16}}\int_{0}^{2\pi}\int_{0}^{\sqrt{17} - \sqrt{1+r^2}} r\, dz\, dr\, d\theta\)

Step 4: Solve the Triple Integral

Now we will solve the triple integral to find the volume: \(\displaystyle V = \int_{0}^{\sqrt{16}}\int_{0}^{2\pi}\int_{0}^{\sqrt{17} - \sqrt{1+r^2}} r\, dz\, dr\, d\theta\) First, we'll integrate with respect to \(z\): \(\displaystyle V = \int_{0}^{\sqrt{16}}\int_{0}^{2\pi} [\frac{rz}{2} \Big\vert_{0}^{\sqrt{17} - \sqrt{1+r^2}}]\, dr\, d\theta\) Which simplifies to: \(\displaystyle V = \int_{0}^{\sqrt{16}}\int_{0}^{2\pi} \frac{r(\sqrt{17} - \sqrt{1+r^2})}{2}\, dr\, d\theta\) Next, we'll integrate with respect to \(r\): \(\displaystyle V = \int_{0}^{2\pi}\Big[\frac{34r}{4} - \frac{r^3}{3}\Big\vert_{0}^{\sqrt{16}}\Big]\, d\theta\) Which simplifies to: \(\displaystyle V = \int_{0}^{2\pi} [34 - \frac{64}{3}]\, d\theta\) Finally, we'll integrate with respect to \(\theta\): \(\displaystyle V = [34\theta - \frac{64\theta}{3}\Big\vert_{0}^{2\pi}\) Simplify and evaluate: \(V = (34(2\pi) - \frac{64(2\pi)}{3}) - (0)\) \(V = \frac{68\pi}{3}\) So, the volume of the solid is \(\displaystyle\frac{68\pi}{3}\).

Key Concepts

Volume Calculation

When dealing with complex shapes in three-dimensional space, finding the volume can be quite challenging. One of the tools mathematicians use is an integral to calculate volumes. Specifically, when a solid has a complicated boundary, as with this hyperboloid shape, integrating can simplify the calculation. This involves setting up and computing a triple integral that corresponds to the volume of the solid. In this exercise, our task was to determine the volume of the solid bounded by a hyperboloid and a plane using cylindrical coordinates. The primary role of this calculation is to integrate over the entire solid region to sum up all the tiny volumes into the total volume. By carefully choosing the right coordinate system — cylindrical coordinates, in this case — we can significantly simplify the computation process and deal effectively with the symmetry inherent in the solid such as circular cross-sections. Using these concepts, one can calculate volumes of complex three-dimensional shapes efficiently.

Triple Integral

A triple integral is an extension of a double integral into three dimensions, used primarily for volume calculation in space. For our problem, the triple integral takes the form:\[ \int\int\int dV \]Here, in cylindrical coordinates, the little volume element, or \(dV\), is expressed as \(r \, dz \, dr \, d\theta \). This conversion of differential elements is crucial as it aligns with how the volume is stacked vertically in a cylindrical shape. The triple integral summarizes how you accumulate the slices (each an infinitesimally small cylinder) across the entire solid. It involves:

• Integrating with respect to \(z\) first, to find the height contribution in each small slice.
• Then with respect to \(r\), summarizing the radial slices.
• Finally integrating over \(\theta\), covering the full circular symmetry.

These integrations allow the volume to be pieced together from basic geometric slices, each extending out until the entire solid is accounted for, providing a pathway from small elements to a complete volume.

Limits of Integration

In integration, specifically triple integration, setting the correct limits of integration is a vital task. These limits define the boundaries over which each variable is evaluated, ensuring that the integral covers every part of the solid.For cylindrical coordinates, the limits of integration are determined by the geometry of the shape:

• **Z limits**: These are defined by the bottom and top surfaces of the solid. Here, the lower limit is the plane \(z = 0\), and the upper limit is \(z = \sqrt{17} - \sqrt{1 + r^2}\), describing the curved top boundary.
• **R limits**: These involve the radius of the foundational circular slice. This particular example considers \(r\) from 0 to \(\sqrt{16}\).
• **Theta limits**: Describing the full rotation around the z-axis, \(\theta\) ranges from 0 to \(2\pi\), completing the circle.

These limits ensure that every part of the solid is considered when calculating the total volume using the triple integral. Correctly setting these limits is fundamental to arriving at the right solution.