Q48P - Introduction to Electrodynamics

Question

Evaluate the following integrals:

(a) $\mathbf{\int }\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{r}\mathbf{·}\mathbf{a}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}{\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{a}\mathbf{)}\mathbf{d\tau }$ , where a is a fixed vector, a is its magnitude.

 

(b) ${\mathbf{\int }}_{\mathbf{v}}{\mathbf{|}\mathbf{r}\mathbf{-}\mathbf{d}\mathbf{|}}^{\mathbf{2}}{\mathbf{\delta }}^{\mathbf{3}}\mathbf{\left(}\mathbf{5}\mathbf{r}\mathbf{\right)}\mathbf{d\tau }$ , where V is a cube of side 2, centered at origin and $\mathbf{b}\mathbf{=}\mathbf{4}\hat{\mathbf{y}}\mathbf{+}\mathbf{3}\hat{\mathbf{z}}$. 

(c) ${\mathbf{\int }}_{\mathbf{v}}{\mathbf{r}}^{\mathbf{4}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{(}\mathbf{r}\mathbf{·}\mathbf{c}\mathbf{)}\mathbf{+}{\mathbf{c}}^{\mathbf{4}}{\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{c}\mathbf{)}\mathbf{d\tau }$ , where is a cube of side 6, about the origin, $\mathbf{c}\mathbf{=}\mathbf{5}\hat{\mathbf{x}}\mathbf{+}\mathbf{3}\hat{\mathbf{y}}\mathbf{+}\mathbf{2}\hat{\mathbf{z}}$ and c is its magnitude.

(d) ${\mathbf{\int }}_{\mathbf{v}}\mathbf{r}\mathbf{·}\mathbf{(}\mathbf{d}\mathbf{-}\mathbf{r}\mathbf{)}{\mathbf{\delta }}^{\mathbf{3}}\mathbf{(}\mathbf{e}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{d\tau }$, where $\mathbf{d}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{e}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}$ , and where v is a sphere of radius  1.5 centered at $(2,2,2)$.

Step-by-Step Solution

Answer

(a) The value of the integral in part (a) is $\int (r^2+r·a+a^2){\delta }^3(r-a)d\tau =3a^2$.

(b) The value of the integral in part (b) is ${\int }_v{|r-d|}^2{\delta }^3\left(5r\right)d\tau =\frac{1}{5}$.

(c) The value of integral in part (c) is ${\int }_v{r}^4+r^2(r·c)+c^4{\delta }^3(r-c)d\tau =0$.

(d) The value of the integral in part (c) is ${\int }_{v}r·(d-r){\delta }^3(e-r)d\tau =-4$ .

Step 1: Describe the given information

The given integrals are $\int ({\mathrm{r}}^2+\mathbf{r}\mathbf{·}\mathbf{a}+{\mathrm{a}}^2){\mathrm{\delta }}^3(\mathbf{r}\mathbf{-}\mathbf{a})\mathrm{d\tau }$, where is a fixed vector, and a is its magnitude. The given integral in (b) is ${\int }_{\mathrm{v}}{|\mathbf{r}\mathbf{-}\mathbf{d}|}^2{\mathrm{\delta }}^3\mathbf{\left(}\mathbf{5}\mathbf{r}\right)\mathrm{d\tau }$ , where v is a cube of side 2, centered at origin, $\mathbf{b}=4\hat{\mathrm{y}}+3\hat{\mathrm{z}}$ . The given integral in (c) is${\int }_{\mathrm{v}}{\mathrm{r}}^4+{\mathrm{r}}^2(\mathbf{r}\mathbf{·}\mathbf{c})+{\mathrm{c}}^4{\mathrm{\delta }}^3(\mathbf{r}\mathbf{-}\mathbf{c})d\tau$, where is a cube of side 6, centered at origin, $\mathbf{c}=5\hat{\mathrm{x}}+3\hat{\mathrm{y}}+2\hat{\mathrm{z}}$ . The given integral in (d) is ${\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})\mathrm{d\tau }$ , where $\mathbf{d}=(1,2,3),\mathbf{ }\mathbf{e}=(3,2,1)$ , and where v is a sphere of radius  1.5 centered at $(2,2,2)$ .

Step 2: Define integral

The integral of a function $\mathit{f}\left(x\right)$, defined as $\mathbf{\int }\mathit{f}\left(x\right)\mathit{d}\mathit{x}$, gives the area bounded by the curve and the x axis.

Step: 3 Find the value of second integral

(a)

The value of the integral in part (b) can be computed using the result, as:

$\delta \left(kx\right)=\frac{1}{\left|k\right|}\delta \left(x\right), \mathrm{as}:$

$$\begin{gathered} \int (r^2+\mathit{r}\mathbf{·}\mathit{a}+a^2){\delta }^3(\mathit{r}\mathbf{-}\mathit{a})d\tau =a^2+a·a+a^2 \\ =a^2+a^2+a^2 \\ =3a^2 \end{gathered}$$

 

Thus the value of the integral in part (b) is $\int (r^2+\mathit{r}\mathbf{·}\mathit{a}+a^2){\delta }^3(\mathit{r}\mathbf{-}\mathit{a})d\tau =3a^2$.

Step: 4 Find the value of first integral.

(b)

The value of the integral in part (a) can be computed using the result $\int f\left(x\right){\delta }^3\left(x-a\right)dx=f\left(a\right)$ , as:

${\int }_v{|\mathit{r}\mathbf{-}\mathit{d}|}^2{\delta }^3\mathbf{\left(}\mathbf{5}\mathit{r}\right)d\tau ={\int }_v{|\mathit{r}\mathbf{-}\mathit{d}|}^2{\delta }^3\mathbf{\left(}\mathbf{5}\mathit{r}\right)\delta \mathbf{\left(}\mathbf{5}\mathit{r}\right)\delta \mathbf{\left(}\mathbf{5}\mathit{r}\right)d\tau$

                                  $$\begin{gathered} =\frac{1}{5^3}{\int }_v{\left|\mathit{r}\mathbf{-}\mathit{b}\right|}^2{\delta }^3\left(r\right)d\tau \\ =\frac{1}{125}{\int }_v{\left|\mathit{r}\mathbf{-}\mathit{b}\right|}^2\delta \left(r\right)d\tau \\ \frac{1}{5^3}{\int }_v{\left|0-\mathit{b}\right|}^2\delta \left(r\right)d\tau \end{gathered}$$

Solve further as, 

 

$$\begin{gathered} {\int }_v{|\mathit{r}\mathbf{-}\mathit{d}|}^2{\delta }^3\mathbf{\left(}\mathbf{5}\mathit{r}\right)d\tau =\frac{{\left|-\mathbf{b}\right|}^2}{5^3}{\int }_v\delta \left(\mathit{r}\right)d\tau \\ =\frac{{\left|-\mathbf{b}\right|}^2}{125}\left(1\right) \\ =\frac{{\left|-4\hat{y}-3\hat{z}\right|}^2}{125} \\ =\frac{{\left(\sqrt{4^2+3^2}\right)}^2}{125} \\ =\frac{1}{5} \end{gathered}$$

 

Thus the value of the integral in part (a) is ${\int }_v{|\mathbf{r}\mathbf{-}\mathbf{d}|}^2{\mathrm{\delta }}^3\mathbf{\left(}\mathbf{5}\mathbf{r}\right)d\tau =\frac{1}{5}$.

Step: 5 Find the value of third integral

(c)

The value of the integral in part (c) can be computed using the fact that the value of Dirac delta function is 0 if the integral ids defined outside its domain.

 

As v is a sphere of radius 6 and $\mathbf{c}=5\hat{\mathrm{x}}+3\hat{\mathrm{y}}+2\hat{\mathrm{z}}$ , where

 

$$\begin{gathered} \left|\mathit{c}\right|=\sqrt{5^2+3^2+2^2} \\ =\sqrt{38} \end{gathered}$$ 

 

Thus, the integral is defined outside the domain of Dirac delta function.

 

Hence ${\int }_{\mathrm{v}}{\mathrm{r}}^4+{\mathrm{r}}^2(\mathbf{r}\mathbf{·}\mathbf{c})+{\mathrm{c}}^4{\mathrm{\delta }}^3(\mathbf{r}\mathbf{-}\mathbf{c})d\tau =0$.

 

The value of integral in part (c) is ${\int }_{\mathrm{v}}{\mathrm{r}}^4+{\mathrm{r}}^2(\mathbf{r}\mathbf{·}\mathbf{c})+{\mathrm{c}}^4{\mathrm{\delta }}^3(\mathbf{r}\mathbf{-}\mathbf{c})d\tau =0$ .

Step: 6 Find the value of fourth integral

(d)

The value of the integral in part (d) is is ${\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})d\tau$, where $\mathbf{d}=(1,2,3),\mathbf{ }\mathbf{e}=(3,2,1)$, and where is a sphere of radius  1.5 centered at $\left(2,2,2\right)$.

 

The value of the integral can be computed using 

 

$$\begin{gathered} {\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})d\tau ={\int }_{\mathrm{v}}\mathbf{e}·\left(\mathbf{d}\mathbf{-}\mathbf{e}\right){\mathrm{\delta }}^3\left(\mathbf{e}\mathbf{-}\mathbf{r}\right)dr \\ =\mathbf{e}·\left(\mathbf{d}\mathbf{-}\mathit{e}\right){\mathrm{\delta }}^3\left(\mathbf{e}\mathbf{-}\mathbf{r}\right)dr \\ =\mathbf{e}·\left(\mathbf{d}\mathbf{-}\mathbf{e}\right)\left(1\right) \\ =\mathbf{e}·\left(\mathbf{d}\mathbf{-}\mathbf{e}\right) \end{gathered}$$

 

The vector $\mathbf{d}=(1,2,3),\mathbf{ }\mathbf{e}=(3,2,1)$ can be written in vector notation as,

 

$$\begin{gathered} \mathbf{d}=\hat{\mathrm{x}}+2\hat{\mathrm{y}}+3\hat{\mathrm{z}}, \\ \mathbf{e}=3\hat{\mathrm{x}}+2\hat{\mathrm{y}}+\hat{\mathrm{z}} \end{gathered}$$

 

Substitute $3\hat{\mathrm{x}}+2\hat{\mathrm{y}}+\hat{\mathrm{z}}$ for $\mathbf{e}\mathbf{,}\mathbf{ }\hat{\mathrm{x}}+2\hat{\mathrm{y}}+3\hat{\mathrm{z}}$  for $\mathbf{d}$ into ${\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})d\tau =\mathbf{e}\mathbf{·}\left(d-e\right)$ .

 

$$\begin{gathered} {\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})d\tau =\left(\begin{array}{c}3\hat{x}+2\hat{y}+\hat{z}\end{array}\right)·\left(\begin{array}{c}\left(\begin{array}{c}\hat{x}+2\hat{y}+3\hat{z}\end{array}\right)\end{array}-\left(\begin{array}{c}3\hat{x}+2\hat{y}+\hat{z}\end{array}\right)\right) \\ =\left(\begin{array}{c}3\hat{x}+2\hat{y}+\hat{z}\end{array}\right)·\left(\begin{array}{c}-\hat{2}x+0\hat{y}+2\hat{z}\end{array}\right) \\ =-6+2 \\ =-4 \end{gathered}$$

 

Thus, the value of the integral in part (c) is ${\int }_{\mathrm{v}}\mathbf{r}·(\mathbf{d}\mathbf{-}\mathbf{r}){\mathrm{\delta }}^3(\mathbf{e}\mathbf{-}\mathbf{r})d\tau =-4$.