6.3.2. Compute the line integral

$$
J=\int_{C}(2 x+4 z) d x+(2 y+2 z) d y-12 z d z
$$

along one turn of the helical line $C$ :

$$
\begin{aligned}
& x=2 \cos 4 t, \\
& y=2 \sin 4 t, \\
& z=6 t, \quad 0 \leq t \leq 2 \pi
\end{aligned}
$$
Solution.

$$
\begin{gathered}
J=\int_{0}^{2 \pi}((4 \cos 4 t+24 t)(-2 \cdot 4 \sin 4 t) d t+ \\
+(4 \sin 4 t+12 t) \cdot(2 \cdot 4 \cos 4 t) d t-12 \cdot 6 \cdot t \cdot 6 d t)= \\
=16 \int_{0}^{2 \pi}(-2 \sin 4 t \cos 4 t-12 t \sin 4 t+2 \sin 4 t \cdot \cos 4 t+6 t \cos 4 t-27 t) d t= \\
=16 \int_{0}^{2 \pi}(-12 t \sin 4 t+6 t \cos 4 t-27 t) d t= \\
=48-4 \int_{0}^{2 \pi} t \sin 4 t d t+2 \int_{0}^{2 \pi} t \cos 4 t d t-9 \int_{0}^{2 \pi} t d t
\end{gathered}
$$

Consider two integrals, which we will solve by parts:

$$
\begin{aligned}
& J_{1}=\int t \sin 4 t d t=-\frac{1}{4} t \cos 4 t+\frac{1}{4} \cdot \frac{1}{4} \int \cos 4 t d(4 t)=-\frac{1}{4} t \cos 4 t+\frac{1}{16} \sin 4 t . \\
& u=t, d u=d t \\
& d v=\sin 4 t d t, v=\frac{1}{4} \int \sin 4 t d(4 t)=-\frac{1}{4} \cos 4 t \\
& J_{2}=\int t \cos 4 t d t=\frac{1}{4} t \sin 4 t-\frac{1}{4} \cdot \frac{1}{4} \int \sin 4 t d(4 t)=\frac{1}{4} t \sin 4 t+\frac{1}{16} \cos 4 t . \\
& u=t, d u=d t \\
& d v=\cos 4 t d t, v=\frac{1}{4} \int \cos 4 t d(4 t)=\frac{1}{4} \sin 4 t
\end{aligned}
$$

We continue to compute the integral \( J \):

$$
\begin{gathered}
J=48-4-\frac{1}{4} t \cos 4 t+\frac{1}{16} \sin 4 t+2 \frac{1}{4} t \sin 4 t+\frac{1}{16} \cos 4 t-\left.\frac{9}{2} t^{2}\right|_{0} ^{2 \pi}= \\
=48 t \cos 4 t-\frac{1}{4} \sin 4 t+\frac{1}{2} t \sin 4 t+\frac{1}{8} \cos 4 t-\left.\frac{9}{2} t^{2}\right|_{0} ^{2 \pi}= \\
=48 \cdot 2 \pi \cos 8 \pi-\frac{1}{4} \sin 8 \pi+\frac{1}{2} \cdot 2 \pi \cdot \sin 8 \pi+\frac{1}{8} \cos 8 \pi-\frac{9}{2} \cdot 4 \pi^{2}-\frac{1}{8} \cos 0= \\
=48 \cdot 2 \pi+\frac{1}{8}-18 \pi^{2}-\frac{1}{8}=96 \pi(1-9 \pi)
\end{gathered}
$$

## 7. ELEMENTS OF FIELD THEORY

## 7.1. Differential Operations