Evaluate the line integral along the curve C.
-$\int _ { C } \frac { x + y + z } { 5 } d s , C$ is the curve $r ( t ) = 4 t i + \left( 6 \cos \frac { 1 } { 2 } t \right) j + \left( 6 \sin \frac { 1 } { 2 } t \right) k , 0 \leq t \leq 2 \pi$
A) $8 \pi ^ { 2 } + 48$
B) $8 \pi$
C) $8 \pi ^ { 2 } + 24$
D) $8 + 24$

Question

Quizplus · Accepted Answer

First, we need to find the curve C in terms of a parameter t. We are given:
$$r ( t ) = 4 t i + \left( 6 \cos \frac { 1 } { 2 } t \right) j + \left( 6 \sin \frac { 1 } { 2 } t \right) k , 0 \leq t \leq 2 \pi$$

Next, we need to find the derivative of r(t) with respect to t, which is:

$$r'(t) = 4i - 3\sin\frac{1}{2}t\; j + 3\cos\frac{1}{2}t\; k$$

The magnitude of $r'(t)$ is:

$$\left \| r'(t) \right \| = \sqrt{(4)^2 + (-3\sin\frac{1}{2}t)^2 + (3\cos\frac{1}{2}t)^2} = 5$$

Therefore, we can set up the line integral as:

$$\int_C \frac{x+y+z}{5}ds = \int_0^{2\pi} \frac{(4t + 6\cos\frac{1}{2}t + 6\sin\frac{1}{2}t)}{5} \left \| r'(t) \right \| dt$$

Simplifying the integrand gives:

$$\frac{4t}{5} + \frac{6}{5}\cos\frac{1}{2}t + \frac{6}{5}\sin\frac{1}{2}t$$

Integrating with respect to t from 0 to $2\pi$ gives:

$$\frac{(4t)^2}{10} \Bigg|_0^{2\pi} + \frac{6}{5} \cdot 2 \sin\frac{1}{2}t \Bigg|_0^{2\pi} - \frac{6}{5} \cdot 2 \cos\frac{1}{2}t \Bigg|_0^{2\pi}$$

Simplifying further gives:

$$\frac{(8\pi)^2}{10} + 0 - \frac{6}{5}(-2) - \frac{6}{5}(2) = 8\pi^2 + 24$$

Therefore, the answer is C.

Evaluate the Line Integral Along the Curve C ∫Cx+y+z5ds,C\int _ { C } \frac { x + y + z } { 5 } d s , C∫C​5x+y+z​ds,C

Evaluate the Line Integral Along the Curve C $\int _ { C } \frac { x + y + z } { 5 } d s , C$