Question

(a) Write a triple integral in spherical coordinates for the volume inside the cone \(z^{2}=x^{2}+y^{2}\) and between the planes \(z=1\) and \(z=2 .\) Evaluate the integral. (b) Do (a) in cylindrical coordinates.

Short answer

The volume in spherical coordinates is \[\frac{7 \pi ( 2 - \sqrt{2} )}{3}\]. The volume in cylindrical coordinates is \[\frac{7 \pi}{3}\].

Step-by-step solution

Understand the Geometry

The cone is described by the equation \[z^2 = x^2 + y^2\]. This can be written as \[z = \rho\] in cylindrical coordinates and \[z = r \cos\theta\] in spherical coordinates. The region is bounded by the planes \[z = 1\] and \[z = 2\].

Step 2(a): Set Up the Triple Integral in Spherical Coordinates

In spherical coordinates, the variables are \((\rho, \theta, \phi)\), where the volume element is \[dV = \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi\]. The bounds for \(\rho\) are from \[1\cos\phi\] to \[2\cos\phi\], for \(\phi\) from \[0\] to \[\frac{\pi}{4}\], and for \(\theta\) from \[0\] to \[2\pi\]. Thus, the triple integral is: \[\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{1/\cos\phi}^{2/\cos\phi} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\].

Step 3(a): Evaluate the Inner Integral

Integrate with respect to \(\rho\): \[\int_{1/\cos\phi}^{2/\cos\phi} \rho^2 \, d\rho\]. This gives: \[\left[ \frac{\rho^3}{3} \right]_{1/\cos\phi}^{2/\cos\phi} = \left( \frac{\left(\frac{2}{\cos\phi}\right)^3}{3} - \frac{\left(\frac{1}{\cos\phi}\right)^3}{3} \right)\].Simplifying this, we get: \[\frac{8}{3 \cos^3 \phi} - \frac{1}{3 \cos^3 \phi} = \frac{7}{3 \cos^3 \phi}\].

Step 4(a): Evaluate the Next Integral

Integrate with respect to \(\phi\): \[\int_{0}^{\frac{\pi}{4}} \frac{7}{3} \sin\phi \, d\phi\]. This becomes: \[\frac{7}{3} \left[-\cos \phi\right]_{0}^{\frac{\pi}{4}}\]. Evaluating this, we get: \[\frac{7}{3} ( -\cos(\frac{\pi}{4}) + \cos(0) ) = \frac{7}{3} ( -\frac{\sqrt{2}}{2} + 1 ) = \frac{7 ( 2 - \sqrt{2} )}{6}\].

Step 5(a): Final Integral with Respect to \(\theta\)

Integrate with respect to \(\theta\): \[\int_{0}^{2\pi} \frac{7 ( 2 - \sqrt{2} )}{6} \, d\theta\]. Since this is a constant: \[\frac{7 ( 2 - \sqrt{2} )}{6} \int_{0}^{2\pi} d\theta = \frac{7 ( 2 - \sqrt{2} )}{6} \times 2\pi = \frac{7 \pi ( 2 - \sqrt{2} )}{3}\].

Step 1(b): Set Up the Integral in Cylindrical Coordinates

In cylindrical coordinates, the variables are \((r, \theta, z)\) with the volume element \[dV = r \, dr \, d\theta \, dz\]. The bounds for \[r\] are from \[0\] to \[z\], for \[z\] are from \[1\] to \[2\], and for \[\theta\] from \[0\] to \[2\pi\]. Thus, the triple integral is: \[\int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{z} r \, dr \, dz \, d\theta\].

Step 2(b): Evaluate the Inner Integral

Integrate with respect to \(r\): \[\int_{0}^{z} r \, dr\]. This gives: \[\left[ \frac{r^2}{2} \right]_{0}^{z} = \frac{z^2}{2}\].

Step 3(b): Evaluate the Next Integral

Integrate with respect to \(z\): \[\int_{1}^{2} \frac{z^2}{2} \, dz\]. This becomes: \[\frac{1}{2} \left[ \frac{z^3}{3} \right]_{1}^{2} = \frac{1}{2} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{1}{2} \times \frac{7}{3} = \frac{7}{6}\].

Step 4(b): Final Integral with Respect to \(\theta\)

Integrate with respect to \(\theta\): \[\int_{0}^{2\pi} \frac{7}{6} \, d\theta\]. Since this is a constant: \[\frac{7}{6} \int_{0}^{2\pi} d\theta = \frac{7}{6} \times 2\pi = \frac{7 \pi}{3}\].

Key concepts

spherical coordinates

Spherical coordinates are a system of curvilinear coordinates that provide an alternative way to specify points in three-dimensional space. These coordinates are defined by three variables: \(\rho\) for the radial distance, \(\theta\) for the azimuthal angle, and \(\phi\) for the polar angle. They are particularly useful in problems with spherical symmetry.

cylindrical coordinates

Cylindrical coordinates offer another method to describe points in 3D space using a combination of linear and angular measurements. In this system, each point is determined by the coordinates \(r\) for the radial distance from the z-axis, \(\theta\) for the azimuthal angle, and \(z\) for the height along the z-axis.

volume integral

A volume integral extends the concept of single integrals to three-dimensional spaces. It computes the total value of a function over a 3D region. In spherical coordinates, the volume element is \(dV = \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi\), while in cylindrical coordinates, it's \(dV = r \, dr \, d\theta \, dz\).

multivariable calculus

Multivariable calculus extends single-variable calculus to functions of several variables. This involves partial derivatives, multiple integrals, and vector calculus. Triple integrals, such as those done in this exercise, are one of its key applications.

geometry of cone

The geometry of the cone described in this exercise involves understanding its representation in different coordinate systems. The cone equation \(z^2 = x^2 + y^2\) can be transformed to \(z = \rho\) in cylindrical coordinates and \(z = r \cos\theta\) in spherical coordinates. This geometric understanding is crucial for setting up the integral bounds correctly.