Problem 64 - 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. The solid inside the sphere \(\rho=1\) and below the cone \(\varphi=\pi / 4\) for \(z \geq 0\)

Step-by-Step Solution

Answer

The volume of the solid inside the sphere ρ=1 and below the cone ϕ=π/4 for z≥0 in spherical coordinates is -π√2/3 + 2π/3.

Step 1: Setting up the volume integral in spherical coordinates

To find the volume of the solid, we can use the following volume integral: $$ V = \int_{V} \rho^2 \sin{\varphi} d\rho d\varphi d\theta $$ Next, we must determine the limits of integration in terms of \(\rho\), \(\varphi\), and \(\theta\). For the radial coordinate \(\rho\), we are considering the solid inside the sphere \(\rho=1\), so the limits will be from 0 to 1. For the polar angle \(\varphi\), all points inside the solid are below the cone with \(\varphi=\pi/4\). Hence, \(\varphi\) will range from 0 to \(\pi/4\). For the azimuthal angle \(\theta\), the solid is restricted to non-negative \(z\)-values, which corresponds to the upper half-space. Therefore, \(\theta\) will range from 0 to \(2\pi\).

Step 2: Computing the volume integral

With the limits of integration set, we can compute the volume integral: $$ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^1 \rho^2 \sin\varphi \, d\rho d\varphi d\theta $$ Let's compute this step by step, starting with the radial integral: $$ \int_0^1 \rho^2 d\rho = \frac{1}{3}\rho^3 \bigg\rvert_0^1 = \frac{1}{3} $$ Next, let's compute the polar integral: $$ \int_0^{\pi/4} \sin\varphi \, d\varphi = -\cos\varphi \bigg\rvert_0^{\pi/4} = -\frac{\sqrt{2}}{2} + 1 $$ Finally, let's compute the azimuthal integral: $$ \int_0^{2\pi} \, d\theta = 2\pi $$ In order to find the volume of the solid, multiply these results together: $$ V = \frac{1}{3} \left(-\frac{\sqrt{2}}{2} + 1\right) (2\pi) = -\frac{\pi\sqrt{2}}{3} + \frac{2\pi}{3} $$ The volume of the solid inside the sphere \(\rho=1\) and below the cone \(\varphi=\pi / 4\) for \(z \geq 0\) is \(-\frac{\pi\sqrt{2}}{3} + \frac{2\pi}{3}\).

Key Concepts

Multiple Integration

Understanding multiple integration is crucial for computing volumes, surface areas, and many other physical quantities. In essence, multiple integration is the extension of single-variable integral calculus to functions of several variables. When solving problems involving the volume of a solid, for instance, we often make use of triple integrals.

A triple integral involves integrating a function of three variables over a three-dimensional region. It is written as \(\int\int\int_{V} f(x, y, z) \,dx\,dy\,dz\), where \(V\) represents the volume over which the integration is to be performed. To perform such an integration, we first find the limits of integration for each variable, which describe the region \(V\), and then calculate the integral, often in an iterated manner — integrating with respect to one variable at a time.

Spherical Coordinates

Spherical coordinates \( (\rho, \varphi, \theta) \) are particularly useful when dealing with problems that have some degree of spherical symmetry, such as the volume of a sphere or a solid bounded by spherical surfaces.

In this system, \(\rho\) is the radial distance from the origin, \(\varphi\) (often referred to as the polar angle) is the angle measured from the positive \(z\)-axis, and \(\theta\) (the azimuthal angle) is the angle from the positive \(x\)-axis in the \(xy\)-plane.

Transitioning to Spherical Coordinates

When converting a volume integral to spherical coordinates, the integrand must include \(\rho^2 \sin\varphi\) to account for the volume element in spherical coordinates, recognized as \(dV = \rho^2 \sin\varphi \,d\rho \,d\varphi \,d\theta\). This ensures the integral accurately represents the actual volume.

Volume of Solids

When faced with finding the volume of a solid, one efficient approach is using integration, particularly when the solid has irregular boundaries or is bounded by surfaces that are best described in non-Cartesian coordinates.

The volume of a solid is essentially the integral of the solid's cross-sectional area as we move along a particular axis or a combination of axes. In spherical coordinates, we consider the 'slices' of the volume in spherical shells, and thus the volume element changes accordingly. By integrating a function that represents the volume of these tiny spherical shells over the limits of our solid, we can calculate the whole volume.

Limits of Integration

Determining the correct limits of integration is key to solving any integration problem, and particularly when computing volumes using triple integrals.

The limits define the region of integration and are based on the geometry of the problem. In spherical coordinates, the limits for \(\rho\), \(\varphi\), and \(\theta\) need to fully encompass the volume of interest.

• \(\rho\) ranges from the origin to the outer bound of the solid in the radial direction.
• \(\varphi\) ranges between the polar angles that define the 'vertical' extent of the solid.
• \(\theta\) typically ranges from 0 to \(2\pi\) to cover the full azimuthal extent, unless the solid has cylindrical symmetry or another form of 'angular' boundary.

Paying close attention to these boundaries ensures that the integral includes the entire volume of the solid in question. Ahence, any section's size should not be underestimated for the limits in integration.