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Chapter 13: Problem 6

Explain why \(\rho^{2} \sin \varphi d \rho d \varphi d \theta\) is the volume of a small "box" in spherical coordinates.

Short Answer

Expert verified
Explain why \(\rho^{2} \sin \varphi d \rho d \varphi d \theta\) is the volume of a small "box" in spherical coordinates. In spherical coordinates, the volume element is expressed as \(\rho^{2} \sin \varphi d \rho d \varphi d \theta\). This expression results from the transformation between Cartesian and spherical coordinate systems, specifically by computing the Jacobian determinant of the transformation matrix. Each term contributes to the volume, with \(\rho^2\) accounting for radial distance, \(\sin \varphi\) in the polar angle, and the differentials (\(d\rho\), \(d\varphi\), and \(d\theta\)) representing small changes in the radial, polar, and azimuthal angles, respectively. Therefore, in a spherical coordinate system, \(\rho^{2} \sin \varphi d \rho d \varphi d \theta\) gives the volume of a small box.

Step by step solution

01

Understand spherical coordinates

To begin, let's recall the definition of spherical coordinates. A spherical coordinate system (Rhophilta) is a three-dimensional coordinate system where a position in space \((x, y, z)\) is defined by three parameters: \(\rho\) (the radial distance),\(\theta\) (the azimuthal angle), and \(\varphi\) (the polar angle). In cartesian coordinates, these can be defined using the following equations: 1. \(x = \rho\sin\varphi\cos\theta\) 2. \(y = \rho\sin\varphi\sin\theta\) 3. \(z = \rho\cos\varphi\)
02

Derive the transformation matrix for Jacobian

Next, let's find the Jacobian, which is the determinant of the matrix formed by the partial derivatives of the transformation from one coordinate system to another. \[J = \begin{bmatrix} \frac{\partial x}{\partial\rho} & \frac{\partial x}{\partial\varphi} & \frac{\partial x}{\partial\theta}\\ \frac{\partial y}{\partial\rho} & \frac{\partial y}{\partial\varphi} & \frac{\partial y}{\partial\theta}\\ \frac{\partial z}{\partial\rho} & \frac{\partial z}{\partial\varphi} & \frac{\partial z}{\partial\theta}\\ \end{bmatrix}\] Now compute the individual derivatives and fill them in: \[J = \begin{bmatrix} \sin\varphi\cos\theta & \rho\cos\varphi\cos\theta & -\rho\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \rho\cos\varphi\sin\theta & \rho\sin\varphi\cos\theta\\ \cos\varphi & -\rho\sin\varphi & 0\\ \end{bmatrix}\]
03

Calculate the Jacobian

Now we compute the determinant of the Jacobian matrix: \[ |\text{J}|= \begin{vmatrix} \sin\varphi\cos\theta & \rho\cos\varphi\cos\theta & -\rho\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \rho\cos\varphi\sin\theta & \rho\sin\varphi\cos\theta\\ \cos\varphi & -\rho\sin\varphi & 0\\ \end{vmatrix}\] \[|\text{J}| = \rho^2 \sin \varphi(\cos^2\theta + \sin^2\theta),\] since \(\cos^2\theta + \sin^2\theta = 1\), we get \[|\text{J}| = \rho^2 \sin \varphi\]
04

Derive the volume of a small box in spherical coordinates

Finally, we will derive the expression for the volume of the small box in spherical coordinates. In Cartesian coordinates, the volume of a small box is given by \(dx\,dy\,dz\). To obtain the equivalent volume expression in spherical coordinates, we multiply the volume element in Cartesian coordinates by the Jacobian of the transformation: \[dV = |\text{J}| \, d\rho \, d\varphi \, d\theta\] Substitute the Jacobian's absolute value we have found earlier: \[dV = (\rho^2 \sin \varphi) \, d\rho \, d\varphi\, d\theta\] Therefore, the volume of a small box in spherical coordinates is given by \(\rho^{2} \sin \varphi \, d\rho \, d\varphi \, d \theta\).

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