91Ó°ÊÓ

To find the gradient of the scalar function \(\varphi(x, y, z)\), we need to compute its partial derivatives with respect to \(x\), \(y\), and \(z\). Gradient: \(\nabla \varphi = \langle \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z}\rangle.\) Given \(\varphi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}\), we can compute the gradient: \(\nabla \varphi = \left\langle \frac{\partial}{\partial x}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial x}\left(\frac{y^2}{2}\right) + \frac{\partial}{\partial x}\left(\frac{z^2}{2}\right), \frac{\partial}{\partial y}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial y}\left(\frac{y^2}{2}\right) + \frac{\partial}{\partial y}\left(\frac{z^2}{2}\right), \frac{\partial}{\partial z}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial z}\left(\frac{y^2}{2}\right) + \frac{\partial}{\partial z}\left(\frac{z^2}{2}\right) \right\rangle.\) After computing the partial derivatives, we get: \(\nabla \varphi = \langle x, y, z \rangle.\) The path \(C\) is given by the parametric equation: \(\mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle.\) Now, let's compute its derivative with respect to \(t\): \(\frac{d\mathbf{r}}{dt} = \langle -\sin t, \cos t, \frac{1}{\pi} \rangle.\)

Now, we will compute the line integral directly using the parametric description of \(C\) and the gradient of the scalar function \(\varphi\). First, let's find the dot product between \(\nabla \varphi\) and \(\frac{d\mathbf{r}}{dt}\) using \(x = \cos t\), \(y = \sin t\), and \(z = \frac{t}{\pi}\). Dot product: \(\nabla \varphi \cdot \frac{d\mathbf{r}}{dt} = \langle x, y, z \rangle \cdot \langle -\sin t, \cos t, \frac{1}{\pi} \rangle = -x \sin t + y \cos t + \frac{z}{\pi}\). Substitute the values for \(x, y,\) and \(z\): \(\nabla \varphi \cdot \frac{d\mathbf{r}}{dt} = -\cos t \cdot \sin t + \sin t \cdot \cos t + \frac{t}{\pi^2}\). Now, let's compute the line integral: \(\int_{C} \nabla \varphi \cdot d\mathbf{r} = \int_{0}^{2\pi} \left(-\cos t \cdot \sin t + \sin t \cdot \cos t + \frac{t}{\pi^2}\right) dt = \int_{0}^{2\pi} \frac{t}{\pi^2} dt.\) Integrate with respect to \(t\): \(\int_{C} \nabla \varphi \cdot d\mathbf{r} = \frac{t^2}{2\pi^2} \Big|_{0}^{2\pi} = \frac{(2\pi)^2}{2\pi^2} - \frac{0}{2\pi^2} = 2.\)

The Fundamental Theorem for line integrals states that: \(\int_{C} \nabla \varphi \cdot d\mathbf{r} = \varphi\left(\mathbf{r}(b)\right) - \varphi\left(\mathbf{r}(a)\right)\). Here, \(a = 0\) and \(b = 2\pi\). We need to compute \(\varphi(\mathbf{r}(0))\) and \(\varphi(\mathbf{r}(2\pi))\). \(\varphi(\mathbf{r}(0)) = \frac{(\cos0)^2 + (\sin0)^2 + \left(\frac{0}{\pi}\right)^2}{2} = \frac{1}{2}\). \(\varphi(\mathbf{r}(2\pi)) = \frac{(\cos2\pi)^2 + (\sin2\pi)^2 + \left(\frac{2\pi}{\pi}\right)^2}{2} = \frac{1+4\pi^2}{2}\). Now, let's compute the line integral using the Fundamental Theorem for line integrals: \(\int_{C} \nabla \varphi \cdot d\mathbf{r} = \varphi\left(\mathbf{r}(2\pi)\right) - \varphi\left(\mathbf{r}(0)\right) = \frac{1+4\pi^2}{2} - \frac{1}{2} = 2\). In both methods, we found the value of the line integral to be \(2\).