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Convert the line integral to an ordinary integral with respect to the parameter and evaluate it. The equation is given as follows:

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

– $C$ 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:

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

5/5 - (16 votes)