91Ó°ÊÓ

The given expressions is

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

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

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

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

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

The Rectangular coordinates are,

$(-5,5,0)$

In Rectangular coordinates:$\mathrm{ÒÏ}=\mathrm{δ}\left(x-x_0\right)\mathrm{δ}\left(y-y_0\right)\mathrm{δ}\left(z-z_0\right)$$\mathrm{ÒÏ}=\mathrm{δ}(x+5)\mathrm{δ}(y-5)\mathrm{δ}\left(z\right)$

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

$\mathrm{ÒÏ}=\frac{\mathrm{δ}(r-5\sqrt{2})\mathrm{δ}(\mathrm{θ}-3\mathrm{Ï€}/4)\mathrm{δ}\left(z\right)}{r}$

And, the spherical coordinates of this particle, is

$$\begin{gathered} r=\sqrt{{(-5)}^2+{\left(5\right)}^2+0^2} \\ =5\sqrt{2} \\ \mathrm{θ}=\frac{3}{4}\mathrm{π}\mathrm{ϕ} \\ ={\cos }^{-1}\left(\frac{0}{5\sqrt{2}}\right) \\ ={\cos }^{-1}\left(0\right)=\frac{\mathrm{π}}{2} \end{gathered}$$

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

$\mathrm{ÒÏ}=\frac{\mathrm{δ}(r-5\sqrt{2})\mathrm{δ}\left(\mathrm{θ}-\frac{3\mathrm{Ï€}}{4}\right)\mathrm{δ}\left(\mathrm{Ï•}-\frac{\mathrm{Ï€}}{2}\right)}{r\sin \left(\mathrm{θ}\right)}$

The Rectangular coordinates are,

$(0,-1,-1)$

In Rectangular coordinates:

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

$$\begin{gathered} \mathrm{ÒÏ}=\frac{\mathrm{δ}\left(r-r_0\right)\mathrm{δ}\left(0-a_0\right)\mathrm{δ}\left(z-z_0\right)}{r} \\ \mathrm{ÒÏ}=\frac{\mathrm{δ}(r-1)\mathrm{δ}\left(0-\frac{3x}{2}\right)\mathrm{δ}(z+1)}{1}\left(\text{Since}{x}^2+y^2=r^2,\tan \mathrm{θ}=-∞\right) \end{gathered}$$

And, the spherical coordinates of this particle, is

$$\begin{gathered} r=\sqrt{{\left(0\right)}^2+{(-1)}^2+{(-1)}^2}=\sqrt{2} \\ \mathrm{θ}=\frac{3\mathrm{π}}{2} \\ \mathrm{ϕ}={\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3\mathrm{π}}{4} \end{gathered}$$

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

$\mathrm{ÒÏ}=\frac{\mathrm{δ}(r-\sqrt{2})\mathrm{δ}\left(\mathrm{θ}-\frac{3\mathrm{Ï€}}{2}\right)\mathrm{δ}\left(\mathrm{Ï•}-\frac{3\mathrm{Ï€}}{4}\right)}{r\sin \left(\mathrm{θ}\right)}$

The Rectangular coordinates are,

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

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

The cylindrical coordinates are,

$$\begin{gathered} \mathrm{ÒÏ}=\frac{\mathrm{δ}\left(r-r_0\right)\mathrm{δ}\left(0-a_0\right)\mathrm{δ}\left(z-z_0\right)}{r} \\ \mathrm{ÒÏ}=\frac{\mathrm{δ}(r-2)\mathrm{δ}(0-\mathrm{Ï€})\mathrm{δ}(z-2\sqrt{3})}{2}\left(\text{Since}{x}^2+y^2=r^2,\tan \mathrm{θ}=-1\right) \end{gathered}$$

The spherical coordinates are,

$$\begin{gathered} r=\sqrt{{(-2)}^2+{\left(0\right)}^2+{\left(2\sqrt{3}\right)}^2}=4 \\ \mathrm{θ}=\mathrm{π} \\ \mathrm{ϕ}={\cos }^{-1}\left(\frac{2\sqrt{3}}{4}\right)=\frac{\mathrm{π}}{6} \end{gathered}$$

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

$\mathrm{ÒÏ}=\frac{\mathrm{δ}(r-4)\mathrm{δ}(\mathrm{θ}-\mathrm{Ï€})\mathrm{δ}\left(\mathrm{Ï•}-\frac{\mathrm{Ï€}}{6}\right)}{r\sin \left(\mathrm{θ}\right)}$

The Rectangular coordinates are,

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

In Rectangular coordinates:

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

The cylindrical coordinates are,

$$\begin{gathered} \mathrm{ÒÏ}=\frac{\mathrm{δ}\left(r-r_0\right)\mathrm{δ}\left(0-{\mathrm{θ}}_0\right)\mathrm{δ}\left(z-z_0\right)}{r} \\ \mathrm{ÒÏ}=\frac{\mathrm{δ}(r-3\sqrt{2})\mathrm{δ}\left(0+\frac{\mathrm{Ï€}}{4}\right)\mathrm{δ}(z+\sqrt{6})}{3\sqrt{2}}\left(\text{Since}{x}^2+y^2=r^2,\tan \mathrm{θ}=-1\right) \end{gathered}$$

The spherical coordinates are,

$$\begin{gathered} r=\sqrt{{\left(3\right)}^2+{(-3)}^2+{(-\sqrt{6})}^2}=2\sqrt{6} \\ \mathrm{θ}=\frac{7\mathrm{π}}{4} \\ \mathrm{ϕ}={\cos }^{-1}\left(\frac{-\sqrt{6}}{2\sqrt{6}}\right)=\frac{2\mathrm{π}}{3} \end{gathered}$$

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

$\mathrm{ÒÏ}=\frac{\mathrm{δ}(r-2\sqrt{6})\mathrm{δ}\left(\mathrm{θ}-\frac{7\mathrm{Ï€}}{4}\right)\mathrm{δ}\left(\mathrm{Ï•}-\frac{2\mathrm{Ï€}}{3}\right)}{r\sin \left(\mathrm{θ}\right)}$