Problem 52 - Calculus Early Transcendentals

Question

Use spherical coordinates to find the volume of the following solids. The solid inside the cone \(z=\left(x^{2}+y^{2}\right)^{1 / 2}\) that lies between the planes \(z=1\) and \(z=2\)

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(3\pi\).

Step 1: Rewrite the cone equation in cylindrical coordinates

We have the equation for the cone given as \(z=\left(x^{2}+y^{2}\right)^{1 / 2}\). To change this to cylindrical coordinates, we substitute \((x, y, z)\) with \((r\cos\theta, r\sin\theta, z)\). We get \(z = r\), which represents the equation of the cone in cylindrical coordinates.

Step 2: Determine the limits of integration

The solid lies inside the cone, so we need to find the limits for \(r\), \(\theta\), and \(z\) that describe the solid. Considering that \(z=r\) since we are inside the cone, we are given that \(1 \leq z \leq 2\). Therefore, we have \(1 \leq r \leq 2\). For the angle \(\theta\), we observe that the solid inside the cone is symmetrical about the \(z\)-axis, so the limits for \(\theta\) are \(0 \leq \theta \leq 2\pi\). Thus, our limits of integration are: - \(1 \leq r \leq 2\) - \(0 \leq \theta \leq 2\pi\) - \(1 \leq z \leq 2\)

Step 3: Integrate the volume element over the limits

The volume \(V\) of the solid can be found using the triple integral of the volume element in cylindrical coordinates: \(V = \int\int\int rdzd\theta dr\) Now, plugging in the limits for the integral, we get: \(V = \int_{1}^{2}\int_{0}^{2\pi}\int_{1}^{2} rdzd\theta dr\)

Step 4: Calculate the integral to find the volume

Evaluating the triple integral, we obtain: \(V = \int_{1}^{2}\int_{0}^{2\pi}\int_{1}^{2} rdz d\theta dr\) \(V = \int_{1}^{2}\int_{0}^{2\pi} [\frac{1}{2}r^2]_{1}^{2} d\theta dr\) \(V = \int_{1}^{2}\int_{0}^{2\pi} (\frac{1}{2}(4) - \frac{1}{2}(1)) d\theta dr\) \(V = \int_{1}^{2}\int_{0}^{2\pi} \frac{3}{2}d\theta dr\) First, integrate over \(\theta\): \(V = \int_{1}^{2} [\frac{3}{2}\theta]_{0}^{2\pi}dr\) \(V =\int_{1}^{2} 3\pi dr\) Now, integrate over \(r\): \(V = [3\pi r]_{1}^{2}\) \(V = 3\pi(2) - 3\pi(1)\) \(V=3\pi\) So, the volume of the solid inside the cone \(z=\left(x^{2}+y^{2}\right)^{1 / 2}\) that lies between the planes \(z=1\) and \(z=2\) is \(3\pi\).

Key Concepts

Spherical Coordinates

Spherical coordinates are an essential tool in solving volume problems for symmetrical shapes. They are defined by three values:

• Radial distance ( ): the distance from the origin to the point.
• Polar angle ({\varphi}): the angle the point makes with a reference plane, typically the xy-plane.
• Azimuthal angle ({\theta}): the rotation around the z-axis.

These coordinates are especially useful for solids like spheres or cones because they naturally fit the geometry. In spherical coordinates, the volume element is expressed as \(dV = \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta\).
This is vital when setting up integrals to calculate volumes for symmetrical objects and simplifies computing because you align your calculations with the object's natural symmetry axes.

Cylindrical Coordinates

Cylindrical coordinates are another coordinate system that combines the simplicity of polar coordinates with the height addition of Cartesian coordinates. They consist of:

• Radial distance ({r}): distance from the z-axis.
• Azimuthal angle ({\theta}): the angle around the z-axis.
• Height ({z}): elevation above the xy-plane.

These coordinates are particularly useful for cylinders and similar shapes.
For example, the cone given by \(z=\sqrt{x^2+y^2}\) becomes \(z = r\) in cylindrical coordinates. This transformation is handy because it simplifies integration when finding the volume of objects bounded within such a shape.
The volume element in cylindrical coordinates is \(dV = r \, dz \, d\theta \, dr\), which is straightforward compared to Cartesian coordinates when dealing with radial symmetry.

Volume Calculation

Calculating volume using triple integrals involves integrating the determined volume element across the specified bounds. First, identify the limits for each coordinate. For the solid in the cone \(z = \sqrt{x^2+y^2}\), we utilize the given bounds:

• {1 \leq r \leq 2} for radial distance inside the cone.
• {0 \leq \theta \leq 2\pi} for full rotation around the z-axis.
• {1 \leq z \leq 2} for the height of the solid between the planes.
  

To compute the volume, you integrate the function representing the volume element over these limits:\[ V = \int_{1}^{2} \int_{0}^{2\pi} \int_{1}^{2} r \, dz \, d\theta \, dr \]Breaking it down, you would first integrate over z, then \(\theta\), and finally r. This results in a simple computation for symmetrical solids.
The final computed volume in this example is \(3\pi\), showcasing how cylindrical coordinates simplify otherwise complex geometric calculations.