Question

Use cylindrical coordinates to find the volume of the following solid regions. The region bounded by the plane \(z=\sqrt{29}\) and the hyperboloid \(z=\sqrt{4+x^{2}+y^{2}}\)

Short answer

Answer: The volume of the solid region bounded by the given plane and hyperboloid is \(50\pi(\sqrt{29} - 3\pi)\).

Step-by-step solution

Convert the plane equation to cylindrical coordinates

The plane is given by the equation, \(z = \sqrt{29}\). As the z-component is unchanged in cylindrical coordinates, the equation will stay the same in cylindrical coordinates: \(z = \sqrt{29}\).

Convert the hyperboloid equation to cylindrical coordinates

The hyperboloid is given by the equation, \(z = \sqrt{4+x^{2}+y^{2}}\). In cylindrical coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\). Therefore, the equation of the hyperboloid becomes \(z = \sqrt{4 + r^{2}(\cos^{2} \theta + \sin^{2} \theta)} = \sqrt{4+r^{2}}\). Step 2: Finding the limits of integration.

Intersection of the hyperboloid and the plane

To find the limits of integration, we need to find where the hyperboloid and the plane intersect. We can do this by equating their equations with respect to the z-coordinate: \(\sqrt{29} = \sqrt{4 + r^{2}}\) Squaring both sides yields: \(29 = 4 + r^{2}\Rightarrow r^{2} = 25 \Rightarrow r = 5\). Thus, the hyperboloid intersects the plane at \(r=5\). Step 3: Setting up and calculating the triple integral.

Set up the triple integral for the volume

We are finding the volume enclosed by the hyperboloid and the plane, so our limits of integration for \(r\) would be from \(0\) to \(5\). The limits for \(\theta\) would be from \(0\) to \(2\pi\) as the surface is symmetric about the z-axis. Finally, the limits for z would be from the equation of the hyperboloid to the equation of the plane, which are \(z = \sqrt{4+r^2}\) and \(z = \sqrt{29}\) respectively. The triple integral in cylindrical coordinates will look like this: \(\int_{0}^{2\pi} \int_{0}^{5} \int_{\sqrt{4+r^2}}^{\sqrt{29}} r \,dz\, dr\, d\theta\)

Evaluate the triple integral

Evaluating the triple integral: \(\int_{0}^{2\pi} \int_{0}^{5} [\sqrt{29} - \sqrt{4+r^2}]r \,dr\, d\theta\) Now, we can find the antiderivative with respect to r and substitute the limits of integration: \(\int_{0}^{2\pi} [\frac{1}{4}(25\sqrt{29}-4r^2 - 4r\sqrt{4+r^2})] \Big|_{0}^{5} \,d\theta\) Evaluating at the limits: \(\int_{0}^{2\pi} [\frac{25}{4}(5\sqrt{29} - 3\pi)] \,d\theta\) Finally, integrating with respect to \(\theta\): \([\frac{25}{4}(5\sqrt{29} - 3\pi) \theta] \Big|_{0}^{2\pi} = \boxed{50\pi(\sqrt{29} - 3\pi)}\) Thus, the volume of the region bounded by the plane and the hyperboloid is equal to \(50\pi(\sqrt{29} - 3\pi)\).

Key concepts

Volume Calculation in Cylindrical Coordinates

When we want to calculate the volume of certain solid regions, especially those bounded by geometric shapes, we often turn to cylindrical coordinates for simplification. By using this coordinate system, we can transform complex shapes into easier forms, thanks to how cylindrical coordinates work with radius \(r\), angle \(\theta\), and height \(z\). In our case, we are looking at a volume beneath a hyperboloid and capped by a plane. Here’s a simple breakdown of how this process works:

* **Step 1:** Identify the boundaries: In this case, they are the plane \(z = \sqrt{29}\) and a hyperboloid \(z = \sqrt{4 + x^2 + y^2}\). In cylindrical coordinates, this becomes \(z = \sqrt{4 + r^2}\).
* **Step 2:** Use the cylindrical form: In cylindrical coordinates, the volume element \(dV\) is expressed as \(r \, dz \, dr \, d\theta\).
* **Step 3:** Set the integral up: Construct a triple integral using the radial symmetry and height differences between the surfaces from the hyperboloid to the plane. Now, integrate over these boundaries to find the volume between them efficiently.

Understanding Triple Integrals

A triple integral allows us to calculate volumes by building up small volume elements over a given region. In cylindrical coordinates, we focus on three variables: radius \(r\), angle \(\theta\), and height \(z\). This setup is particularly suited for problems involving circular symmetry, where cylindrical coordinates shine.

* **Function Setup:** In our scenario, the triple integral takes the form:
\[ \int_{0}^{2\pi} \int_{0}^{5} \int_{\sqrt{4+r^2}}^{\sqrt{29}} r \, dz \, dr \, d\theta \]
* **Order of Integration:** Here, we integrate first with \(z\), then \(r\), and finally \(\theta\). This order is determined by the geometry of the problem, allowing us to smoothly integrate outward from the axis of symmetry.

Triple integrals essentially 'add up' infinitely tiny volumes defined by function \(f(r,\theta,z)\), as scaled by the volume element \(r\,dz\,dr\,d\theta\). When worked through, this provides an exact volume for the region bounded by our given surfaces.

Determining Limits of Integration

In any integral setup, establishing the correct limits of integration is crucial. These limits define the edges or bounds of our region of interest within which we want to calculate the volume. In cylindrical coordinates, limits vary for \(r\), \(\theta\), and \(z\). Here's how it works practically in this problem:

* **For \(r\):** The radial component, \(r\), ranges from the center out to where the hyperboloid meets the plane. The critical intersection was calculated as \(r = 5\) where \(\sqrt{29} = \sqrt{4 + r^2}\).
* **For \(\theta\):** Due to the rotational symmetry around the z-axis, \(\theta\) spans a full circle from \(0\) to \(2\pi\), covering all angles.
* **For \(z\):** Heights \(z\) vary from the hyperboloid \(z = \sqrt{4+r^2}\) up to the plane \(z = \sqrt{29}\) above.

Getting the limits correct ensures our integration accurately captures the entire volume beneath the plane and above the hyperboloid. This strategy avoids underestimating or overestimating the volume by precisely defining the bounds of the integration region.