Problem 78 - Calculus Early Transcendentals

Question

General volume formulas Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Spherical cap Find the volume of the cap of a sphere of radius \(R\) with thickness \(h\).

Step-by-Step Solution

Answer

Answer: The volume of a spherical cap with radius R and thickness h is given by the formula: \(\displaystyle V = \frac{4\pi hR^2}{3}\).

Step 1: Review of spherical coordinates and recall spherical cap

Spherical coordinates consist of three coordinates: \((r, \theta, \phi)\). In Cartesian coordinates, these are related as: $x = r \sin\phi \cos\theta \\ y = r \sin\phi \sin\theta \\ z = r \cos\phi$ We are given a spherical cap that is a part of the sphere with radius \(R\) and thickness \(h\).

Step 2: Setting the limits of integration

Since we are dealing with a sphere whose radius is \(R\), we know it's bounded by \(r\) from 0 to \(R\). For the cap part, we need to take care of its thickness \(h\). We can set the limits for the variable \(\phi\) as follows: The cap is just touching the plane \(z = R - h,\) which means for \(\phi\), \(r\cos\phi = R - h\). Since \(r = R,\) we get \(\cos\phi = \frac{R-h}{R}\) and \(\phi = \arccos(\frac{R-h}{R}).\) Therefore, the limits for \(\phi\) will be from \(\arccos(\frac{R-h}{R})\) to \(\frac{\pi}{2}\). And for \(\theta,\) we can use the full circle \(2\pi\) as its limit. Thus, the limits for \(\theta\) will be from \(0\) to \(2\pi\).

Step 3: Setting up the triple integral

Now we can set up the triple integral to find the volume of the spherical cap. The volume element in the spherical coordinate system is given by \(dV = r^2\sin\phi dr d\theta d\phi.\) Using this and the defined limits of integration, we get: \(V = \int_{0}^{2\pi} \int_{\arccos(\frac{R-h}{R})}^{\frac{\pi}{2}} \int_{0}^{R} r^2\sin\phi dr d\theta d\phi\)

Step 4: Evaluate the triple integral

Now we have to evaluate the triple integral that we set up: \(V = \int_{0}^{2\pi} d\theta \int_{\arccos(\frac{R-h}{R})}^{\frac{\pi}{2}} d\phi \int_{0}^{R} r^2\sin\phi dr\) First, let's integrate with respect to \(r\): \(V= \int_{0}^{2\pi} d\theta \int_{\arccos(\frac{R-h}{R})}^{\frac{\pi}{2}} d\phi \left[\frac{r^3\sin\phi}{3}\right]_{0}^{R}\) Which simplifies to: \(V = \frac{R^3}{3} \int_{0}^{2\pi} d\theta \int_{\arccos(\frac{R-h}{R})}^{\frac{\pi}{2}} \sin\phi d\phi\) Now, we integrate with respect to \(\phi\): \(V = \frac{R^3}{3} \int_{0}^{2\pi} d\theta \left[-\cos\phi\right]_{\arccos(\frac{R-h}{R})}^{\frac{\pi}{2}}\) Which simplifies to: \(V = \frac{R^3}{3} \int_{0}^{2\pi} d\theta (1-\frac{R-h}{R})\) \(V = \frac{hR^2}{3} \int_{0}^{2\pi} d\theta\) Finally, we integrate with respect to \(\theta\): \(V = \frac{hR^2}{3} \left[\theta\right]_{0}^{2\pi}\) Which simplifies to: \(V = \frac{4\pi hR^2}{3}\) Hence, the volume of the spherical cap with radius \(R\) and thickness \(h\) is given by: \(V = \frac{4\pi hR^2}{3}\)

Key Concepts

Spherical Coordinates

Spherical coordinates are a system used to define the position of points in a three-dimensional space using three numbers, often represented by \( (r, \theta, \phi) \). Unlike Cartesian coordinates which use \( (x, y, z) \), spherical coordinates are based on the distance from the origin and two angles.

• \( r \) is the radial distance from the origin to the point.
• \( \theta \) is the azimuthal angle, representing rotation around the \( z \) axis, ranging from \( 0 \) to \( 2\pi \).
• \( \phi \) is the polar angle, depicting the angle down from the \( z \) axis, generally from \( 0 \) to \( \pi \).

This coordinate system is particularly useful when dealing with spherical objects or problems that exhibit radial symmetry.When converting from spherical to Cartesian coordinates, the relationships used are:

• \( x = r \sin\phi \cos\theta \)
• \( y = r \sin\phi \sin\theta \)
• \( z = r \cos\phi \)

Spherical coordinates simplify the integration of volume and surface integrals for objects with spherical symmetry. They are instrumental in simplifying the calculations, especially for spheres, spheroids, and other similar structures.

Spherical Cap

A spherical cap is a shape derived from cutting a sphere with a plane, resulting in a dome-like section. It's very similar to a lid lifted from the top of a sphere.If you imagine a sphere as a full ball, the spherical cap will sit on top after slicing the ball horizontally. The key characteristics defining a spherical cap include:

• The radius of the sphere \( R \).
• The height or thickness of the cap \( h \), which is the distance between the slicing plane and the highest point of the spherical cap.

The spherical cap's volume can be calculated by understanding its geometric requirements. The thickness and the radius are crucial in determining the limit settings during integral setup.In the example of a spherical cap of radius \( R \), touching a plane at \( z = R - h \), integration limits are defined with respect to these characteristics. By utilizing spherical coordinates, we simplify the representation and calculation of such three-dimensional sections.

Volume of Solids

Finding the volume of a solid object can vary in complexity depending on the shape of the solid. When tackling problems involving spheres or parts of spheres, like a spherical cap, using integration through spherical coordinates offers an efficient method.In our case study of a spherical cap, the integration is set up across three dimensions:

• The radial coordinate \( r \), ranging from \( 0 \) to \( R \).
• The polar angle \( \phi \), adjusted for the cap from \( \arccos\left(\frac{R-h}{R}\right) \) to \( \frac{\pi}{2} \).
• The azimuthal angle \( \theta \), which encompasses the full circle from 0 to \( 2\pi \).

The integral of the volume element \( r^2 \sin\phi \, dr \, d\theta \, d\phi \) is evaluated over these limits to determine the volume of the spherical cap.The calculated volume for the spherical cap, using definite integration, turns out to be \( \frac{4\pi hR^2}{3} \), highlighting the power of integration in providing detailed and accurate volume calculations for geometrically complex shapes like a spherical cap.