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Chapter 14: Problem 28

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. $$\varphi(x, y)=\left(x^{2}+y^{2}\right) / 2 ; C: \mathbf{r}(t)=\langle\sin t, \cos t\rangle, \text { for } 0 \leq t \leq \pi$$

Short Answer

Expert verified
Question: Evaluate the line integral $\int_{C} \nabla \varphi \cdot d\mathbf{r}$ for the gradient vector field $\nabla \varphi(x, y) = \langle x, y \rangle$ and the curve $C$ given by $\mathbf{r}(t) = \langle \sin t, \cos t \rangle$, for $0 \leq t \leq \pi$. Use both the direct evaluation method and the Fundamental Theorem for Line Integrals. Answer: The line integral $\int_{C} \nabla \varphi \cdot d\mathbf{r}$ is equal to $0$ for the given gradient vector field and curve $C$.

Step by step solution

01

Find the gradient of the scalar function \(\varphi(x,y)\)

The gradient of a scalar function \(\varphi(x,y)\) is given by \(\nabla \varphi = \left(\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}\right)\). So, let's find the partial derivatives with respect to \(x\) and \(y\) for the given function \(\varphi(x, y)=\left(x^{2}+y^{2}\right) / 2\). The partial derivative with respect to \(x\) is \(\frac{\partial \varphi}{\partial x} = x\), and with respect to \(y\) is \(\frac{\partial \varphi}{\partial y} = y\). Thus, the gradient of \(\varphi(x,y)\) is $$\nabla \varphi = \langle x, y \rangle$$
02

Parameterize the curve C using the vector function \(r(t)\)

The curve \(C\) is already given as a parametric representation using the vector function \(r(t)\). So, we have $$\mathbf{r}(t) = \langle \sin t, \cos t \rangle , \text { for } 0 \leq t \leq \pi$$
03

Calculate \(d\mathbf{r}\) and \(\nabla \varphi \cdot d\mathbf{r}\)

Find the total differential of \(\mathbf{r}(t)\): $$d\mathbf{r} = \frac{d\mathbf{r}}{dt}dt = \langle \cos t, - \sin t \rangle dt$$ Now, substitute the parametric representation of the curve \(C\) into the gradient \(\nabla \varphi\): $$\nabla \varphi (\mathbf{r}(t)) = \langle \sin t, \cos t \rangle$$ Now we can find the dot product of \(\nabla \varphi (\mathbf{r}(t))\) and \(d\mathbf{r}\): $$\nabla \varphi (\mathbf{r}(t))\cdot d\mathbf{r} = \langle \sin t, \cos t \rangle \cdot \langle \cos t, -\sin t \rangle dt$$ $$= (\sin t\cos t - \cos t\sin t) dt = 0$$
04

Evaluate the line integral directly

Finally, we can evaluate the line integral: $$\int_{C} \nabla \varphi \cdot d\mathbf{r} = \int_{0}^{\pi} 0 dt = 0$$ #b. Evaluation of the Line Integral Using the Fundamental Theorem for Line Integrals#
05

Verify that the vector field is conservative

For a scalar function \(\varphi(x, y)\), its gradient vector field \(\nabla \varphi\) is always conservative. We can therefore use the Fundamental Theorem for Line Integrals.
06

Evaluate the potential function along the curve endpoints

We need to evaluate the potential function, \(\varphi(x,y)\), at the endpoints of the curve \(C\). The endpoints are given by \(\mathbf{r}(0) = \langle \sin(0), \cos(0) \rangle = \langle 0, 1 \rangle\) and \(\mathbf{r}(\pi) = \langle \sin(\pi), \cos(\pi) \rangle = \langle 0, -1 \rangle\). Now, evaluate \(\varphi(x, y)\) at these points: $$\varphi(0, 1) = \frac{0^2+1^2}{2} = \frac{1}{2}$$ $$\varphi(0, -1) = \frac{0^2+(-1)^2}{2} = \frac{1}{2}$$
07

Apply the Fundamental Theorem for Line Integrals

The Fundamental Theorem for Line Integrals states that for a conservative vector field \(\mathbf{F} = \nabla \varphi\), the line integral over curve \(C\) is the difference in the potential function at the endpoints: $$\int_{C} \nabla \varphi \cdot d\mathbf{r} = \varphi(B) - \varphi(A)$$ with \(A\) and \(B\) being the initial and final points of the curve, respectively. For our problem, we have: $$\int_{C} \nabla \varphi \cdot d\mathbf{r} = \varphi(0, -1) - \varphi(0, 1) = \frac{1}{2} - \frac{1}{2} = 0$$ Both methods yield the same result, showing that the line integral of the given gradient vector field \(\nabla \varphi\) over the curve \(C\) is \(0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient
Understanding the gradient is crucial for solving line integrals involving scalar functions. The gradient of a scalar function, denoted as \( abla \varphi \), is a vector field. It points in the direction where the function increases most rapidly and its magnitude indicates the rate of increase.

For a function \( \varphi(x, y) = \frac{x^2 + y^2}{2} \), the gradient can be calculated by finding the partial derivatives of \( \varphi \) with respect to \( x \) and \( y \).
  • The partial derivative with respect to \( x \) is \( \frac{\partial \varphi}{\partial x} = x \).
  • The partial derivative with respect to \( y \) is \( \frac{\partial \varphi}{\partial y} = y \).
Therefore, the gradient is represented as \( abla \varphi = \langle x, y \rangle \). This vector helps us understand how the line integral behaves along the curve.
Parametric Representation
A parametric representation simplifies the calculation of line integrals by expressing curves through parameters, usually \( t \).

For the curve \( C \), given by \( \mathbf{r}(t) = \langle \sin t, \cos t \rangle \), the range \( 0 \leq t \leq \pi \) describes a path from the top to the bottom of the circle.
  • The derivative \( d\mathbf{r} = \langle \cos t, -\sin t \rangle dt \) gives us a vector tangent to the curve at any point.
By substituting this parametric form into our gradient, we perform the dot product \( abla \varphi \cdot d\mathbf{r} \) to evaluate the line integral directly. This process bridges the connection between the curve and the scalar function.
Fundamental Theorem for Line Integrals
The Fundamental Theorem for Line Integrals simplifies evaluating integrals over conservative vector fields. It states that the line integral of a gradient \( abla \varphi \) over a curve \( C \) is equal to the change in the potential function \( \varphi \) between the endpoints.

In our exercise, \( \mathbf{r}(0) = \langle 0, 1 \rangle \) and \( \mathbf{r}(\pi) = \langle 0, -1 \rangle \) are the endpoints.
  • At \( A = \mathbf{r}(0) \), \( \varphi(0, 1) = \frac{1}{2} \).
  • At \( B = \mathbf{r}(\pi) \), \( \varphi(0, -1) = \frac{1}{2}\).
The theorem gives \( \int_{C} abla \varphi \cdot d\mathbf{r} = \varphi(B) - \varphi(A) = 0 \). This reasoning is not only efficient but highlights the nature of conservative fields, where path independence means the integral depends solely on the endpoints.

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Most popular questions from this chapter

Prove that the radial field \(\mathbf{F}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(p\) is a real number, is conservative on any region not containing the origin. For what values of \(p\) is \(\mathbf{F}\) conservative on a region that contains the origin?

Suppose \(y=f(x)\) is a continuous and positive function on \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. a. Show that \(S\) is described parametrically by \(\mathbf{r}(u, v)=\langle u, f(u) \cos v, f(u) \sin v\rangle,\) for \(a \leq u \leq b, 0 \leq v \leq 2 \pi\) b. Find an integral that gives the surface area of \(S\) c. Apply the result of part (b) to find the area of the surface generated with \(f(x)=x^{3},\) for \(1 \leq x \leq 2\) d. Apply the result of part (b) to find the area of the surface generated with \(f(x)=\left(25-x^{2}\right)^{1 / 2},\) for \(3 \leq x \leq 4\).

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T\) (the Laplacian of \(T\)). Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-\sqrt{x^{2}+y^{2}+z^{2}}}$$

Let \(f\) be differentiable and positive on the interval \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. Use Theorem 14.12 to show that the area of \(S\) (as given in Section 6.6 ) is $$ \int_{a}^{b} 2 \pi f(x) \sqrt{1+f^{\prime}(x)^{2}} d x $$.

Recall the Product Rule of Theorem \(14.11: \nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) a. Integrate both sides of this identity over a solid region \(D\) with a closed boundary \(S\) and use the Divergence Theorem to prove an integration by parts rule: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions. c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\) where \(D\) is the cube in the first octant cut by the planes \(x=1\) \(y=1,\) and \(z=1\)
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