Problem 12 - University Calculus: Early Transcendentals

Question

Let \(D\) be the region bounded below by the cone \(z=\sqrt{x^{2}+y^{2}}\) and above by the paraboloid \(z=2-x^{2}-y^{2} .\) Set up the triple integrals in cylindrical coordinates that give the volume of \(D\) using the following orders of integration. a. \(d z d r d \theta\) b. \(d r d z d \theta\) c. \(d \theta d z d r\)

Step-by-Step Solution

Answer

a. \(\int_0^{2\pi} \int_0^1 \int_r^{2-r^2} r \, dz \, dr \, d\theta\). b. \(\int_0^{2\pi} \int_0^2 \int_0^{\sqrt{2-z}} r \, dr \, dz \, d\theta\). c. \(\int_0^1 \int_r^{2-r^2} \int_0^{2\pi} r \, d\theta \, dz \, dr\).

Step 1: Understanding the region D

The region D is the volume enclosed between the cone and the paraboloid. The cone is given by \(z = \sqrt{x^2 + y^2}\), while the paraboloid is described by \(z = 2 - x^2 - y^2\). We need to express these surfaces in cylindrical coordinates.

Step 2: Express surfaces in cylindrical coordinates

In cylindrical coordinates, we have \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The cone equation becomes \(z = r\). The paraboloid becomes \(z = 2 - r^2\).

Step 3: Determine bounds in cylindrical coordinates

From the cone \(z = r\) and the paraboloid \(z = 2 - r^2\), we find the intersection by setting the equations equal: \(r = 2 - r^2\), leading to \(r^2 + r - 2 = 0\). Solving gives \(r = 1\) (valid range for radius in the first quadrant, as the other solution is negative).

Step 4: Set up integral for order \(dz\ dr\ d\theta\)

For this integration order: - \(z\) varies from \(r\) to \(2-r^2\).- \(r\) varies from \(0\) to \(1\).- \(\theta\) varies from \(0\) to \(2\pi\).The integral is: \[ \int_0^{2\pi} \int_0^1 \int_r^{2-r^2} r \, dz \, dr \, d\theta \]

Step 5: Set up integral for order \(dr\ dz\ d\theta\)

For this order: - \(r\) varies from \(0\) to \(\sqrt{2-z}\).- \(z\) varies from \(r\) to \(2\).- \(\theta\) varies from \(0\) to \(2\pi\).The integral is: \[ \int_0^{2\pi} \int_0^2 \int_0^{\sqrt{2-z}} r \, dr \, dz \, d\theta \]

Step 6: Set up integral for order \(d\theta\ dz\ dr\)

For this order: - \(\theta\) varies from \(0\) to \(2\pi\).- \(z\) varies from \(r\) to \(2-r^2\).- \(r\) varies from \(0\) to \(1\).The integral is: \[ \int_0^{1} \int_{r}^{2-r^2} \int_0^{2\pi} r \, d\theta \, dz \, dr \]

Key Concepts

Triple Integrals

Triple integrals are a powerful tool in calculus used to compute volumes of three-dimensional regions among other applications. In essence, a triple integral extends the concept of an integral from one or two dimensions into three dimensions. These integrals can calculate quantities that depend on three variables, such as mass, volume, or charge distribution in a 3D region.

When setting up a triple integral, it often involves specifying the order of integration. This order defines which variable you integrate first, second, and third. The common orders of integration include \(dz \, dr \, d\theta\), \(dr \, dz \, d\theta\), and \(d\theta \, dz \, dr\). Each order will depend on the nature of the bounds and surfaces that define the region you're integrating over. Using cylindrical coordinates can simplify the process when dealing with symmetrical problems like the one with a cone and paraboloid.

Ultimately, mastering triple integrals involves understanding both the geometric nature of the problem and the mathematical setup of the integral.

Volume Calculation

In mathematical problems involving triple integrals, calculating the volume of a 3D object is a common task. Volume calculation using integrals relies on evaluating how small sections of volume can be summed up to find the total. This is crucial when you're working with objects bounded by surfaces like cones and paraboloids.

For the volume between surfaces described in cylindrical coordinates, the volume is calculated by setting up the triple integral with appropriate bounds. The bounds are derived by considering the intersection and limits of the surfaces in the cylindrical coordinates. In our problem, the volume calculation requires integrating a function involving the variable \(r\), which represents the radial distance from the origin.

The general form for calculating volume in cylindrical coordinates is represented as follows: \[ \int_{\text{range of } \theta} \int_{\text{range of } r} \int_{\text{range of } z} r \, dz \, dr \, d\theta \] The factor \(r\) accounts for the change from Cartesian coordinates into cylindrical coordinates, ensuring the volume is calculated correctly.

Cone and Paraboloid Intersection

Understanding the intersection between a cone and a paraboloid is essential in setting the bounds for triple integrals. These intersections define the limits of the region you are interested in. In our exercise, the cone is represented by the equation \(z = \sqrt{x^2 + y^2}\), and the paraboloid by \(z = 2 - x^2 - y^2\).

To convert these into cylindrical coordinates, use the transformations: \(x = r\cos\theta\) and \(y = r\sin\theta\). This changes the cone's equation to \(z = r\) and the paraboloid's to \(z = 2 - r^2\). The intersection occurs when both equations give the same \(z\) value, leading to \(r = 2 - r^2\). Solving this gives the valid range of \(r\) for the region.

Finding intersections between such surfaces is about understanding their geometric relationship. The cone and paraboloid intersect along a circle in a plane parallel to the base of the cone. This circle defines the maximum \(r\) value, which is crucial for setting up the bounds in the integrals. In our case, it simplifies to a radius of 1, indicating the limit of integration for \(r\). Understanding these intersections and transformations allows for proper setup of the integral that accurately calculates the volume enclosed between the surfaces.