Question

Evaluate the line integral \(\int_{C} \nabla \varphi \cdot d \mathbf{r}\) for the following functions \(\varphi\) and oriented curves \(C\) in two ways. a. Use a parametric description of \(C\) to evaluate the integral directly. b. Use the Fundamental Theorem for line integrals. $$\begin{aligned} &\varphi(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right) / 2 ; C: \mathbf{r}(t)=\langle\cos t, \sin t, t / \pi\rangle, \text { for }\\\ &0 \leq t \leq 2 \pi \end{aligned}$$

Short answer

In summary, we calculated the line integral of a scalar function $\varphi$ along the path $C$ in two different ways: using the parametric description of the path and using the Fundamental Theorem for line integrals. In both cases, we found the value of the line integral to be $2$. This demonstrates that the two methods give the same result for this particular problem.

Step-by-step solution

Find the gradient of the scalar function \(\varphi\) and the derivative of the path \(C\)

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

Compute the line integral using a parametric description of \(C\)

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 \(t\): \(\int_{C} \nabla \varphi \cdot d\mathbf{r} = \frac{t^2}{2\pi^2} \Big|_{0}^{2\pi} = \frac{(2\pi)^2}{2\pi^2} - \frac{0}{2\pi^2} = 2.\)

Compute the line integral using the Fundamental Theorem for line integrals

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 \(2\).

Key concepts

Gradient Vector Field

A gradient vector field is derived from a scalar function and provides insight into its behavior. For a given scalar function, the gradient of the function points in the direction of the maximum rate of increase of the function. It is represented as a vector field. To compute the gradient of a scalar function \( \varphi(x, y, z) \), you take partial derivatives with respect to each independent variable. For example, with \( \varphi(x, y, z) = \frac{x^2 + y^2 + z^2}{2} \), the gradient \( abla \varphi \) results in \( \langle x, y, z \rangle \). This process illustrates how a single scalar function can give rise to a vector field that describes its changing values across space.

Fundamental Theorem for Line Integrals

The Fundamental Theorem for Line Integrals is similar to the Fundamental Theorem of Calculus, but for vector fields. It links the line integral of a gradient vector field with the values of the scalar function at the endpoints of the path. This theorem states that if a vector field is the gradient of a scalar function, then the line integral is simply the difference between the function values at the curve's endpoints. Specifically, for a curve \( C \) from point \( a \) to point \( b \), it is expressed as:\[ \int_{C} abla \varphi \cdot d\mathbf{r} = \varphi(\mathbf{r}(b)) - \varphi(\mathbf{r}(a)) \] This method simplifies the calculation of line integrals significantly, as demonstrated by evaluating the endpoints \( \mathbf{r}(0) \) and \( \mathbf{r}(2\pi) \) in the original problem, which yielded the same result as the parametric approach.

Parametric Equations

Parametric equations provide a way to describe a path or curve in space using one or more parameters. It is a valuable tool for evaluating line integrals as it allows expressions in terms of a single variable. For instance, the curve \( C \) given in the problem is represented as \( \mathbf{r}(t) = \langle\cos t, \sin t, \frac{t}{\pi}\rangle \). Each component of \( \mathbf{r}(t) \) is described in terms of the parameter \( t \), which varies over an interval, such as from \( 0 \) to \( 2\pi \) in this problem. Using parametric equations makes it straightforward to compute derivatives and integrals along a path, setting the foundation for evaluating line integrals directly.

Dot Product

The dot product is a key operation in vector calculus which combines two vectors to produce a scalar. It is fundamentally important when evaluating line integrals because it measures how aligned two vectors are. In the context of line integrals, you take the dot product of the gradient vector field \( abla \varphi \) with the derivative of the position vector, \( \frac{d\mathbf{r}}{dt} \). The result gives you the integrand for the line integral. The formula for the dot product is:\[ \langle a, b, c \rangle \cdot \langle d, e, f \rangle = ad + be + cf \] In our specific example, this results in simplifying expressions such as \( -\cos t \cdot \sin t + \sin t \cdot \cos t + \frac{t}{\pi^2} \), which provides the expression needed to integrate over the given parameter interval.

Integral Calculus

Integral calculus is a branch of mathematics dealing with the concept of integration, one of the fundamental operations in calculus. It is the process of finding the integral of a function, which can be used to determine areas under curves, among other applications. In the context of line integrals, integration is performed on a function over a curve, combining calculus and vector analysis. The integral of a function along a curve sums up values of the function from start to end of the path. By integrating the dot product of the gradient vector with the curve's derivative, we compute the line integral over the interval, transforming it from a geometric path into a scalar quantity, as shown where \( \int_{0}^{2\pi} \frac{t}{\pi^2} dt \) results in \( 2 \). Understanding this operation simplifies the evaluation of functions over a given domain.