Question

Let \(R\) be a region in a plane that has a unit normal vector \(\mathbf{n}=\langle a, b, c\rangle\) and boundary \(C .\) Let \(\mathbf{F}=\langle b z, c x, a y\rangle\) a. Show that \(\nabla \times \mathbf{F}=\mathbf{n}\) b. Use Stokes' Theorem to show that $$\operatorname{area} \text { of } R=\oint_{C} \mathbf{F} \cdot d \mathbf{r}$$ c. Consider the curve \(C\) given by \(\mathbf{r}=\langle 5 \sin t, 13 \cos t, 12 \sin t\rangle\) for \(0 \leq t \leq 2 \pi .\) Prove that \(C\) lies in a plane by showing that \(\mathbf{r} \times \mathbf{r}^{\prime}\) is constant for all \(t\) d. Use part (b) to find the area of the region enclosed by \(C\) in part (c). (Hint: Find the unit normal vector that is consistent with the orientation of \(C\).)

Short answer

Explain your answer. A: Yes, the curve lies in a plane as it was proven that the cross product of the position vector r(t) and its derivative r'(t) is constant for all values of t. This implies that the curve lies within a plane.

Step-by-step solution

Part a: Finding the curl of the vector field (Proof)

First, let's find the curl of the given vector field \(\mathbf{F} = \langle bz, cx, ay \rangle\). The curl of a vector field is given by: $$\nabla \times \mathbf{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \times \left\langle bz, cx, ay \right\rangle$$ Calculating the components of the curl, we get: $$\nabla \times \mathbf{F} = \left\langle \frac{\partial ay}{\partial y} - \frac{\partial cx}{\partial z}, \frac{\partial bz}{\partial z} - \frac{\partial ay}{\partial x}, \frac{\partial cx}{\partial x} - \frac{\partial bz}{\partial y} \right\rangle = \langle a, b, c \rangle$$ This is equal to the unit normal vector \(\mathbf{n}\), as required.

Part b: Applying Stokes' Theorem (Proof)

Now we will use Stokes' Theorem to show that the area of the region \(R\) can be expressed as the line integral of \(\mathbf{F}\) over its boundary \(C\). Stokes' Theorem states that: $$\iint_{R} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}$$ Since we found that \(\nabla \times \mathbf{F} = \mathbf{n}\) in part (a), we can write the left side of Stokes' Theorem as: $$\iint_{R} \mathbf{n} \cdot d\mathbf{S}$$ Now, recall that this integral represents the area of region \(R\). Therefore, we have proven that: $$\operatorname{area} \text { of } R = \oint_C \mathbf{F} \cdot d\mathbf{r}$$

Part c: Proving that the curve lies in a plane (Proof)

We are given the curve \(C\) defined by the position vector \(\mathbf{r}(t) = \langle 5 \sin t, 13 \cos t, 12 \sin t \rangle\) for \(0 \leq t \leq 2 \pi\). To prove that the curve lies in a plane, we need to show that the cross product \(\mathbf{r} \times \mathbf{r}^\prime\) is constant for all \(t\). First, let's find \(\mathbf{r}^\prime(t)\). Taking the derivative of \(\mathbf{r}\) with respect to \(t\), we get: $$\mathbf{r}^\prime(t) = \left\langle \frac{d}{dt} (5 \sin t), \frac{d}{dt} (13 \cos t), \frac{d}{dt} (12 \sin t) \right\rangle = \langle 5 \cos t, -13 \sin t, 12 \cos t \rangle$$ Now, we will compute the cross product of \(\mathbf{r}\) and \(\mathbf{r}^\prime\): $$\mathbf{r} \times \mathbf{r}^\prime = \left\langle (13 \cos t)(12 \cos t) - (-13 \sin t)(12 \sin t), (5 \sin t)(12 \cos t) - (12 \sin t)(5 \cos t), (5 \sin t)(-13 \sin t) - (13 \cos t)(5 \cos t) \right\rangle = \langle 0, 0, -144 \rangle$$ Since \(\mathbf{r} \times \mathbf{r}^\prime\) is a constant vector, we can deduce that the curve lies in a plane.

Part d: Finding the area of the region enclosed by the curve (Calculation)

Finally, we will find the area of the region enclosed by the curve \(C\), using the integral expression derived in part (b). First, let's find the unit normal vector for the given curve. Since \({\bf r} \times {\bf r}' = \langle 0,0,-144\rangle\), the normal vector to the plane is \(\mathbf{n} = \langle 0, 0, -1\rangle\). To calculate the line integral, we need to find \(\mathbf{F} \cdot d\mathbf{r}\). Substituting the given \(\mathbf{F} = \langle bz, cx, ay \rangle\) and the parametric equations for the curve \(C\) into this expression, we have: $$\mathbf{F} \cdot d\mathbf{r} = (0)(5 \cos t) + (-1)(13 \cos t)(-13 \sin t) + (0)(12 \cos t) = 169 \sin t \cos t$$ Now we can evaluate the line integral: $$\operatorname{area} \text { of } R = \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2\pi} 169 \sin t \cos t \, dt$$ To solve this integral, we can use the double-angle formula for the sine function: $$\sin 2\theta = 2\sin\theta\cos\theta$$ Let \(\theta = t\), so we have: $$\sin 2t = 2\sin t \cos t$$ Now the integral becomes: $$\operatorname{area} \text { of } R = \int_{0}^{2\pi} 169 \left(\frac{1}{2} \sin 2t\right) \, dt = 169 \left[-\frac{1}{4} \cos 2t\right]_0^{2\pi} = 169 (0 - 0) = 0$$ Interestingly, the area of the region enclosed by the curve is zero. This is because the curve \(C\) traces a closed trajectory in a plane, with the starting and ending points coinciding.

Key concepts

Vector Field

A vector field is essentially a function that assigns a vector to every point in space. Imagine a weather map where each point in a region is marked with an arrow indicating wind direction and strength — that's a real-world vector field! In mathematical terms, if you're given vector field \(\mathbf{F}\), then for a point \( (x, y, z) \), the value of \(\mathbf{F}\) at that point could look something like \(\mathbf{F}(x, y, z) = \langle P(x, y, z), Q(x, y, z), R(x, y, z) \rangle\). Here, \(P\), \(Q\), and \(R\) are the components of the vector field that depend on the coordinates in the space. Vector fields are key players in physics and engineering, helping to visualize and analyze dynamic systems. They provide insightful information on how a quantity changes over a region.

Curl of a Vector Field

The curl of a vector field is a measure of the rotation of the field around a point. Picture stirring a cup of coffee with a spoon and noticing the swirling motion — that's akin to the curl at work in a fluid. Mathematically, the curl is a vector itself and tells us how much the vector field 'twists' around a point.For a vector field \(\mathbf{F} = \langle P, Q, R \rangle\), the curl is given as:\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \]This operation is like the equivalent of differentiating for vector fields, detecting any swirling patterns the field may have around a point.

Line Integral

Line integrals are a vital tool in calculus, particularly when dealing with vector fields. They allow you to integrate a function along a curve, essentially measuring how much a vector field pushes or pulls as you travel along its path. It's like walking along a path with the wind at your back and calculating the total force you feel along the way.To compute the line integral of a vector field \(\mathbf{F}\) along a curve \(C\), we evaluate:\[ \oint_C \mathbf{F} \cdot d\mathbf{r} \]Here, \(d\mathbf{r}\) represents an infinitesimal segment of the curve, and the dot product \(\mathbf{F} \cdot d\mathbf{r}\) results in how much the field aligns with these segments. This integral sums these interactions over the entire curve \(C\), offering insights into the work done by the field along the path.

Unit Normal Vector

A unit normal vector is a perpendicular vector to a surface with a length of one. This is crucial for defining the orientation of surfaces or planes in space, helping in computing quantities that involve direction, such as fluxes through a surface.In mathematical terms, if a vector \(\mathbf{n}\) is normal to a surface, it means:

• The dot product of \(\mathbf{n}\) and any tangent vector to the surface is zero.
• \(\mathbf{n}\) has unit length, i.e., \(||\mathbf{n}|| = 1\).

This unit normal vector is vital for using Stokes' Theorem, where it contributes to transforming surface integrals into curve integrals. By specifying the orientation of the surface, it helps in calculating how a vector field crosses it.