\[ \int_C (y -\ z) \, ds \]

**– $C$ is the helix path $r(t) = < 4 \cos t, 4 \sin t, t > \hspace{0.3in} for\ 0 \leq t \leq 2 \pi$.**

This question aims to find the **integration** of the **line integral** after converting it to an **ordinary integral** according to the **given parameters.**

The question is based on the concept of **line integral. Line integral** is the integral where the function of the **line** is integrated along the given **curve.** Line integral is also known as **path integral, curve integral,** and sometimes **curvilinear integral.**

## Expert Answer

The given **limits** of the function are as follows:

\[ r(t) = (4 \cos t)i + (4 \sin t)j + (t)k \hspace{0.5in} on\ 0 \leq t \leq 2 \pi \]

\[ x = 4 \cos t \]

\[ y = 4 \sin t \]

\[ z = t \]

Taking the **derivatives** of all the above **limits** with respect to $t$ on both sides as:

\[ dfrac{dx} {dt} = \dfrac{d} {dt} 4 \cos t\]

\[ dx = -4 \sin t dt \]

\[ dfrac{dy} {dt} = \dfrac{d} {dt} 4 \sin t\]

\[ dy = 4 \cos t dt \]

\[ dz = dt \]

The $r'(t)$ will become:

\[ r'(t) = < -4 \sin t, 4 \cos t, 1 > \]

Calculating the magnitude of the $r'(t)$ as:

\[ r'(t) = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2 + 1^2} \]

\[ r'(t) = \sqrt{ (16 \sin^2 t) + (16 \cos^2 t) + 1} \]

\[ r'(t) = \sqrt{ 17 (\sin^2 t + \cos^2 t)} \]

\[ r'(t) = \sqrt{17} \]

Now we can find the **ordinary integral** of the given **line integral** as:

\[ \int_C (y -\ z) \, ds = \int_{0}^{2 \pi} (y -\ z) r'(t) \, dt \]

\[ \int_{0}^{2 \pi} (y -\ z) r'(t) \, dt \]

Substituting the values, we get:

\[ \int_{0}^{2 \pi} (4 \sin t -\ t) \sqrt{17} \, dt \]

Solving the **integral,** we get:

\[ = \sqrt{17} \Big[ -4 \cos t -\ \dfrac{t^2} {2} \Big]_{0}^{2 \pi} \]

\[ = \sqrt{17} \Big[ -4 – 2 \pi^2 + 4 \Big] \]

\[ = -2 \pi^2 \sqrt{17} \]

## Numerical Result

The **ordinary integral** of the **line integral** given is calculated to be:

\[ \int_C (y -\ z) \, ds = -2 \pi^2 \sqrt{17} \hspace{0.5in} on\ 0 \leq t \leq 2 \pi \]

## Example

Calculate the **integral** of the given **curve** over $0 \leq x \leq 2\pi$.

\[ f(x) = x^2 + \dfrac{x}{2} \]

The **integral** can be calculated by simply using the **limits** of the given **curve** and solving over the **integrated equation.**

\[ \int_ {0}^ {2\pi} f(x) \, dx = \int_ {0}^ {2\pi} x^2 + \dfrac{x}{2} \, dx \]

\[ \int_ {0}^ {2\pi} f(x) \, dx = \Big[ \dfrac{x^3} {3} + \dfrac{x^2} {4} \Big]_{0}^{2\pi} \]

\[ \int_ {0}^ {2\pi} f(x) \, dx = \Big[ \dfrac{(2\pi)^3}{3} + \dfrac{(2\pi)^2}{4} \Big] -\ 0 \]

\[ \int_ {0}^ {2\pi} f(x) \, dx = \pi^2 \Big( 1 + \dfrac{8 \pi}{3} \Big) \]

Simplifying the values, we get:

\[ \int_ {0}^ {2\pi} f(x) \, dx = 92.55 \]