Evaluate the line integral Integral from Upper C to font size decreased by 10 gradient phi times d Bold r for the following function phi and oriented curve C ​(a) using a parametric description of C and evaluating the integral​ directly, and ​(b) using the Fundamental Theorem for line integrals.
phi​(x,y,z)equalsStartFraction x squared plus y squared plus z squared Over 2 EndFraction
​;       ​ C: r​(t)equalsleft angle cosine t comma sine t comma StartFraction t Over pi EndFraction right angle
​, for StartFraction 3 pi Over 4 EndFraction
less than or equalstless than or equalsStartFraction 5 pi Over 4 EndFraction

To evaluate the line integral 

\[
\int_C \nabla \phi \cdot d\mathbf{r}
\]

for the given scalar function \(\phi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}\) along the curve \(C: \mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle\) over the interval \(t \in \left[ \frac{3\pi}{4}, \frac{5\pi}{4} \right]\), we can use both (a) a parametric approach, and (b) the Fundamental Theorem for line integrals.

### Step (a): Direct Evaluation Using Parametric Description

1. **Compute \(\nabla \phi\):** 

   \[
   \phi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}
   \]

   The gradient \(\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right)\) is:

   \[
   \frac{\partial \phi}{\partial x} = x, \quad \frac{\partial \phi}{\partial y} = y, \quad \frac{\partial \phi}{\partial z} = z
   \]

   Thus, 

   \[
   \nabla \phi = \langle x, y, z \rangle
   \]

2. **Parametrize \(\nabla \phi\) along \(C\):**

   Since \(\mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle\), we substitute \(x = \cos t\), \(y = \sin t\), and \(z = \frac{t}{\pi}\) into \(\nabla \phi\):

   \[
   \nabla \phi(\mathbf{r}(t)) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle
   \]

3. **Compute \(d\mathbf{r}\):**

   Differentiate \(\mathbf{r}(t)\) with respect to \(t\):

   \[
   \frac{d\mathbf{r}}{dt} = \langle -\sin t, \cos t, \frac{1}{\pi} \rangle
   \]

   So, \(d\mathbf{r} = \langle -\sin t, \cos t, \frac{1}{\pi} \rangle \, dt\).

4. **Compute \(\nabla \phi \cdot d\mathbf{r}\):**

   Now, calculate the dot product \(\nabla \phi(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}\):

   \[
   \nabla \phi \cdot \frac{d\mathbf{r}}{dt} = \langle \cos t, \sin t, \frac{t}{\pi} \rangle \cdot \langle -\sin t, \cos t, \frac{1}{\pi} \rangle
   \]

   Expanding this dot product:

   \[
   = (\cos t)(-\sin t) + (\sin t)(\cos t) + \frac{t}{\pi} \cdot \frac{1}{\pi}
   \]

   Simplify each term:

   \[
   = -\cos t \sin t + \cos t \sin t + \frac{t}{\pi^2}
   \]

   The terms \(-\cos t \sin t\) and \(\cos t \sin t\) cancel each other out, leaving:

   \[
   = \frac{t}{\pi^2}
   \]

5. **Integrate over \(t\):**

   Now, integrate from \(t = \frac{3\pi}{4}\) to \(t = \frac{5\pi}{4}\):

   \[
   \int_C \nabla \phi \cdot d\mathbf{r} = \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{t}{\pi^2} \, dt
   \]

   This simplifies to:

   \[
   = \frac{1}{\pi^2} \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} t \, dt
   \]

   Evaluating the integral:

   \[
   = \frac{1}{\pi^2} \cdot \left[ \frac{t^2}{2} \right]_{\frac{3\pi}{4}}^{\frac{5\pi}{4}}
   \]

   Substituting the limits:

   \[
   = \frac{1}{\pi^2} \cdot \left( \frac{(5\pi/4)^2}{2} - \frac{(3\pi/4)^2}{2} \right)
   \]

   \[
   = \frac{1}{\pi^2} \cdot \frac{1}{2} \left( \frac{25\pi^2}{16} - \frac{9\pi^2}{16} \right)
   \]

   Simplify further:

   \[
   = \frac{1}{\pi^2} \cdot \frac{1}{2} \cdot \frac{16\pi^2}{16} \left( 25 - 9 \right)
   \]



Evaluate the line integral Integral from Upper C to font size decreased by 10 gradient phi times d Bold r for the following function phi and oriented curve C ​(a) using a parametric description of C and evaluating the integral​ directly, and ​(b) using the Fundamental Theorem for line integrals. phi​(x,y,z)equalsStartFraction x squared plus y squared plus z squared Over 2 EndFraction ;       ​ C: r​(t)equalsleft angle cosine t comma sine t comma StartFraction t Over pi EndFraction right angle ​, for StartFraction 3 pi Over 4 EndFraction less than or equalstless than or equalsStartFraction 5 pi Over 4 EndFraction

Learn how to evaluate a line integral using both a parametric approach and the Fundamental Theorem for Line Integrals. This solution involves calculating the gradient of the scalar function \(\phi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}\) and integrating over the curve \(C: \mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle\) in the interval \(t \in \left[ \frac{3\pi}{4}, \frac{5\pi}{4} \right]\). Suitable for undergraduate-level calculus students.

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