Question

Write a formula in rectangular coordinates, in cylindrical coordinates, and in spherical coordinates for the density of a unit point charge or mass at the point with the given rectangular coordinates:

(a) $\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}$

(b) $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{)}$

(c)$\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{2}\sqrt{\mathbf{3}}\mathbf{)}$

(d) $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\sqrt{\mathbf{6}}\mathbf{)}$

Short answer

The solutions of statements are

(a) $\rho =\frac{\delta (r-5\sqrt{2})\delta \left(\theta -\frac{3\pi }{4}\right)\delta \left(\varphi -\frac{\pi }{2}\right)}{r\sin \left(\theta \right)}$

(b) $\rho =\frac{\delta (r-\sqrt{2})\delta \left(\theta -\frac{3\pi }{2}\right)\delta \left(\varphi -\frac{3\pi }{4}\right)}{r\sin \left(\theta \right)}$

(c) $\rho =\frac{\delta (r-4)\delta (\theta -\pi )\delta \left(\varphi -\frac{\pi }{6}\right)}{r\sin \left(\theta \right)}$

(d) $\rho =\frac{\delta (r-2\sqrt{6})\delta \left(\theta -\frac{7\pi }{4}\right)\delta \left(\varphi -\frac{2\pi }{3}\right)}{r\sin \left(\theta \right)}$

Step-by-step solution

Given information

The given expressions is

(a)$(-5,5,0)$

(b)$(0,-1,-1)$

(c)$(-2,0,2\sqrt{3})$

(d)$(3,-3,-\sqrt{6})$

Definition of Integration By Parts

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

Solve the given statement at (-5,5,0)

The Rectangular coordinates are,

$(-5,5,0)$

In Rectangular coordinates:$\rho =\delta \left(x-x_0\right)\delta \left(y-y_0\right)\delta \left(z-z_0\right)$$\rho =\delta (x+5)\delta (y-5)\delta \left(z\right)$

The cylindrical coordinates are,Thus, the particle is charge "or mass" density in the cylindrical coordinates is given by

$\rho =\frac{\delta (r-5\sqrt{2})\delta (\theta -3\pi /4)\delta \left(z\right)}{r}$

And, the spherical coordinates of this particle, is

$$\begin{gathered} r=\sqrt{{(-5)}^2+{\left(5\right)}^2+0^2} \\ =5\sqrt{2} \\ \theta =\frac{3}{4}\pi \varphi \\ ={\cos }^{-1}\left(\frac{0}{5\sqrt{2}}\right) \\ ={\cos }^{-1}\left(0\right)=\frac{\pi }{2} \end{gathered}$$

Thus, the density of this point particle in the spherical coordinates is given by

$\rho =\frac{\delta (r-5\sqrt{2})\delta \left(\theta -\frac{3\pi }{4}\right)\delta \left(\varphi -\frac{\pi }{2}\right)}{r\sin \left(\theta \right)}$

Solve the given statement at (0,-1,-1)

The Rectangular coordinates are,

$(0,-1,-1)$

In Rectangular coordinates:

$$\begin{gathered} \rho =\delta \left(x-x_0\right)\delta \left(y-y_0\right)\delta \left(z-z_0\right) \\ \rho =\delta \left(x\right)\delta (y+1)\delta (z+1) \end{gathered}$$

The cylindrical coordinates are,

$$\begin{gathered} \rho =\frac{\delta \left(r-r_0\right)\delta \left(0-a_0\right)\delta \left(z-z_0\right)}{r} \\ \rho =\frac{\delta (r-1)\delta \left(0-\frac{3x}{2}\right)\delta (z+1)}{1}\left(\text{Since}{x}^2+y^2=r^2,\tan \theta =-\infty \right) \end{gathered}$$

And, the spherical coordinates of this particle, is

$$\begin{gathered} r=\sqrt{{\left(0\right)}^2+{(-1)}^2+{(-1)}^2}=\sqrt{2} \\ \theta =\frac{3\pi }{2} \\ \varphi ={\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3\pi }{4} \end{gathered}$$

And hence, the density of this point particle in the spherical coordinates is given by

$\rho =\frac{\delta (r-\sqrt{2})\delta \left(\theta -\frac{3\pi }{2}\right)\delta \left(\varphi -\frac{3\pi }{4}\right)}{r\sin \left(\theta \right)}$

Solve the given statement at (-2,0,23)

The Rectangular coordinates are,

$(-2,0,2\sqrt{3})$

In Rectangular coordinates:$\rho =\delta \left(x-x_0\right)\delta \left(y-y_0\right)\delta \left(z-z_0\right)$$\rho =\delta (x+2)\delta \left(y\right)\delta (z-2\sqrt{3})$

The cylindrical coordinates are,

$$\begin{gathered} \rho =\frac{\delta \left(r-r_0\right)\delta \left(0-a_0\right)\delta \left(z-z_0\right)}{r} \\ \rho =\frac{\delta (r-2)\delta (0-\pi )\delta (z-2\sqrt{3})}{2}\left(\text{Since}{x}^2+y^2=r^2,\tan \theta =-1\right) \end{gathered}$$

The spherical coordinates are,

$$\begin{gathered} r=\sqrt{{(-2)}^2+{\left(0\right)}^2+{\left(2\sqrt{3}\right)}^2}=4 \\ \theta =\pi \\ \varphi ={\cos }^{-1}\left(\frac{2\sqrt{3}}{4}\right)=\frac{\pi }{6} \end{gathered}$$

Thus, the density of this point particle in the spherical coordinates is given by

$\rho =\frac{\delta (r-4)\delta (\theta -\pi )\delta \left(\varphi -\frac{\pi }{6}\right)}{r\sin \left(\theta \right)}$

Solve the given statement at (3,-3,-6)

The Rectangular coordinates are,

$(3,-3,-\sqrt{6})$

In Rectangular coordinates:

$$\begin{gathered} \rho =\delta \left(x-x_0\right)\delta \left(y-y_0\right)\delta \left(z-z_0\right) \\ \rho =\delta (x-3)\delta (y+3)\delta (z+\sqrt{6}) \end{gathered}$$

The cylindrical coordinates are,

$$\begin{gathered} \rho =\frac{\delta \left(r-r_0\right)\delta \left(0-{\theta }_0\right)\delta \left(z-z_0\right)}{r} \\ \rho =\frac{\delta (r-3\sqrt{2})\delta \left(0+\frac{\pi }{4}\right)\delta (z+\sqrt{6})}{3\sqrt{2}}\left(\text{Since}{x}^2+y^2=r^2,\tan \theta =-1\right) \end{gathered}$$

The spherical coordinates are,

$$\begin{gathered} r=\sqrt{{\left(3\right)}^2+{(-3)}^2+{(-\sqrt{6})}^2}=2\sqrt{6} \\ \theta =\frac{7\pi }{4} \\ \varphi ={\cos }^{-1}\left(\frac{-\sqrt{6}}{2\sqrt{6}}\right)=\frac{2\pi }{3} \end{gathered}$$

Thus, the density of this point particle in the spherical coordinates is given by

$\rho =\frac{\delta (r-2\sqrt{6})\delta \left(\theta -\frac{7\pi }{4}\right)\delta \left(\varphi -\frac{2\pi }{3}\right)}{r\sin \left(\theta \right)}$