Chapter 3, Problem 45P

(a)

To determine

The integral ${\displaystyle {\int }_{L}Q\cdot dl}$.

Explanation of Solution

Given:

The slant height of the cone $\left(l\right)$ is $2\text{m}$.

The cone angle $\left(\theta \right)$ is $30°$.

The vector field $\left(Q\right)$ is $\frac{\sqrt{x^2+y^2+z^2}}{\sqrt{x^2+y^2}}\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]$.

Calculation:

Calculate the radius of the cone $\left(r\right)$ using the relation.

  $\begin{array}{c}r=l\sin 30°\\ =\left(2\text{m}\right)\sin 30°\\ =1\text{m}\end{array}$

Calculate the height $\left(h\right)$ of the cone using the relation.

  $\begin{array}{c}h=l\cos 30°\\ =\left(2\text{m}\right)\times \frac{\sqrt{3}}{2}\\ =\sqrt{3}\text{m}\end{array}$

Write the equation of the cone using the relation.

  $\begin{array}{l}{x}^2+y^2={\left(\frac{r}{h}\right)}^2{z}^2\\ x^2+y^2={\left(\frac{1}{\sqrt{3}}\right)}^2{z}^2\\ \frac{z^2}{x^2+y^2}=3\end{array}$

Substitute the value in the vector field $\left(Q\right)$ using the relation.

  $\begin{array}{c}Q=\frac{\sqrt{x^2+y^2+z^2}}{\sqrt{x^2+y^2}}\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]\\ =\sqrt{1+z^2}\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]\\ =\sqrt{1+3}\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]\\ =2\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]\end{array}$

Calculate the integral $\left({\displaystyle {\int }_{L}Q.dl}\right)$ using the relation.

  $\begin{array}{c}{\displaystyle {\int }_{L}Q\cdot d\text{l}}={\displaystyle {\int }_{L}2\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]\cdot \left(dx{a}_x+dy{a}_y+dz{a}_z\right)}\\ =2{\displaystyle \int \left(x-y\right)dx+\left(x+y\right)dy}\\ =2{\displaystyle \iint \left[\frac{\partial }{\partial x}\left(x-y\right)-\frac{\partial }{\partial y}\left(x+y\right)\right]}ds\\ =2{\displaystyle \iint 2ds}\end{array}$

  $\begin{array}{c}=4{\displaystyle \iint ds}\\ =4\left(\pi r^2\right)\\ =4\left(\pi {\left(1\right)}^2\right)\\ =4\pi \end{array}$

Thus, the value of the integral is $\underset{\_}{4\pi }$.

(b)

To determine

The integral ${\displaystyle {\int }_{s_1}\left(\nabla \times Q\right)}\cdot dS$.

Explanation of Solution

Calculation:

Calculate the curl $\left(\nabla \times Q\right)$ using the relation.

  $\begin{array}{c}\nabla \times Q=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 2\left(x-y\right)& 2\left(x+y\right)& 0\end{array}\right|\\ =4a_z\end{array}$

Calculate the integral $\left({\displaystyle {\int }_{s_1}\left(\nabla \times Q\right)}ds\right)$ using the relation.

  $\begin{array}{c}{\displaystyle {\int }_{s_1}\left(\nabla \times Q\right)}ds={\displaystyle \iint \left(4a_x\right)ds}\\ ={\displaystyle \iint \left(4a_x\right)\cdot dxdy{a}_z}\\ ={\displaystyle \iint 4dxdy}\\ =4\left(2\pi {\left(1\text{}\text{m}\right)}^2\right)\\ =8\pi \end{array}$

Thus, the value of the integral is $\underset{\_}{8\pi }$.

(c)

To determine

The integral ${\displaystyle {\int }_{s_2}\left(\nabla \times Q\right)}\cdot dS$.

Explanation of Solution

Calculate the integral using the relation.

$\begin{array}{c}{\displaystyle {\int }_{s_2}\left(\nabla \times Q\right)}ds={\displaystyle \iint \left(4a_z\right)}\cdot \left(-a_{z}dxdy\right)\\ =-{\displaystyle \iint 4dxdy}\\ =-4\left(\pi rl\right)\\ =-4\left(\pi \times 1\times 2\right)\\ =-8\pi \end{array}$

Thus, the value of the integral is $\underset{\_}{-8\pi }$.

(d)

To determine

The integral ${\displaystyle {\int }_{s_1}Q\cdot dS}$.

Explanation of Solution

Calculate the integral $\left({\displaystyle {\int }_{s_1}Q\cdot dS}\right)$ using the relation.

  $\begin{array}{c}{\displaystyle {\int }_{s_1}Q\cdot dS}={\displaystyle \int \nabla \cdot \left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]}dV\\ ={\displaystyle \int \left[\frac{\partial }{\partial x}\left(x+y\right)-\frac{\partial }{\partial y}\left(x-y\right)\right]dV}\\ ={\displaystyle \int 2dV}\\ =2\times \frac{2}{3}\times \pi \times {\left(1\right)}^3\\ =4.186\end{array}$

Thus, the integral ${\displaystyle {\int }_{s_1}Q\cdot dS}$ is $\underset{\_}{4.186}$.

(e)

To determine

The integral ${\displaystyle {\int }_{s_2}Q\cdot dS}$.

Explanation of Solution

Calculate the integral $\left({\displaystyle {\int }_{s_2}Q\cdot dS}\right)$ using the relation.

  $\begin{array}{c}{\displaystyle {\int }_{s_2}Q\cdot dS}={\displaystyle \int \nabla \cdot \left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]dV}\\ =2{\displaystyle \int \left[\frac{\partial }{\partial x}\left(x+y\right)-\frac{\partial }{\partial y}\left(x-y\right)\right]dV}\\ =2{\displaystyle \int dV}\\ =2\times \left(\frac{1}{3}\times \pi \times {\left(1\right)}^2\times \sqrt{3}\right)\\ =3.6276\end{array}$

Thus, the integral ${\displaystyle {\int }_{s_2}Q\cdot dS}$ is $\underset{\_}{3.6276}$.

(f)

To determine

The integral ${\displaystyle {\int }_v\nabla \cdot QdV}$.

Explanation of Solution

Calculate the integral $\left({\displaystyle {\int }_v\nabla \cdot QdV}\right)$ using the relation.

  $\begin{array}{c}{\displaystyle {\int }_v\nabla \cdot QdV}={\displaystyle \underset{v}{\int }\nabla \cdot 2\left[\left(x-y\right)a_x+\left(x+y\right)a_y\right]dV}\\ =2{\displaystyle \underset{v}{\int }dV}\\ =2\left[\frac{2}{3}\pi {\left(1\right)}^2+\frac{1}{3}\pi {\left(1\right)}^2\sqrt{3}\right]\\ =7.8164\end{array}$

Thus, the integral $\left({\displaystyle {\int }_v\nabla \cdot QdV}\right)$ is $\underset{\_}{7.8164}$.

The result in part (a) is the line integral of the base of cone, whereas the result in part (f) includes the volume integral about the combine shape i.e. hemi-sphere and the cone.