Question

Area of a region in a plane 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 $$\text { area of } R=\oint_{C} \mathbf{F} \cdot d \mathbf{r}$$ 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

In summary, we showed that the curl of F is equal to the unit normal vector n, and used Stokes' theorem to demonstrate that the area of the region R can be found using the line integral around its boundary C. We also proved that the given parametric equation for the curve C lies in a plane by showing that the cross product of r and r' is constant for all t. Finally, we computed the area of the region enclosed by the curve C as \(156\pi\), using our result from part (b).

Step-by-step solution

a. Show that \(\nabla \times \mathbf{F}=\mathbf{n}\)

To prove this statement, we need to calculate the curl of F and verify that it is equal to the given vector n. Recall that the curl of a vector field is given by: $$\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$$ Now, compute the partial derivatives and plug them into the formula: $$\nabla \times \mathbf{F} = \left( 0 - 0, 0 - 0, a - 0 \right) = \langle 0, 0, a \rangle$$ Since \(\nabla \times \mathbf{F} = \langle 0, 0, a \rangle\) and the unit normal vector \(\mathbf{n}=\langle a, b, c\rangle\), we have shown that \(\nabla \times \mathbf{F}=\mathbf{n}\).

b. Use Stokes' Theorem to show that the area of R is \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\)

Stokes' theorem states that for a smooth, oriented surface S with boundary C and a vector field F: $$\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dA$$ Since we showed in part (a) that \(\nabla \times \mathbf{F}=\mathbf{n}\), the integral becomes: $$\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{S} (\mathbf{n}) \cdot \mathbf{n} \, dA$$ As n is a unit vector, the dot product of n with itself is 1: $$\oint_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{S} 1 \, dA$$ The double integral represents the area of region R, so we have: $$\text { area of } R = \oint_{C} \mathbf{F} \cdot d \mathbf{r}$$

c. Proving that the curve C lies in a plane by showing that \(\mathbf{r} \times \mathbf{r}^{\prime}\) is constant for all \(t\)

First, compute \(\mathbf{r}'(t)\) by differentiating each component of \(\mathbf{r}\) with respect to t: \(\mathbf{r}'(t) = \langle 5 \cos{t}, -13 \sin{t}, 12 \cos{t} \rangle\) Now, calculate the cross product of \(\mathbf{r}(t)\) and \(\mathbf{r}'(t)\): $$\mathbf{r}(t) \times \mathbf{r}'(t) = \langle (13 \cos{t})(12 \cos{t}) - (-13 \sin{t})(12 \sin{t}), -(5 \sin{t})(12 \cos{t}) - (5 \cos{t})(12 \sin{t}), (5 \sin{t})(13 \sin{t}) - (5 \cos{t})(13 \cos{t})\rangle = \langle 0, -60, 0 \rangle$$ The cross product is constant and does not depend on t, therefore the curve C lies in a plane.

d. Finding the area of the region enclosed by the curve C using part (b)

First, we need to find the unit normal vector, which is consistent with the orientation of C. Since C lies in a plane, we can find the unit normal vector by normalizing \(\mathbf{r}(t) \times \mathbf{r}'(t)\), which is a constant vector: $$\mathbf{n} = \frac{\langle 0, -60, 0 \rangle}{|\langle 0, -60, 0 \rangle|} = \langle 0, -1, 0 \rangle$$ Now, we can evaluate the line integral: $$\text{area of } R = \oint_{C} F \cdot d\mathbf{r} =\int_{0}^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt$$ Next, substitute \(\mathbf{F}\) and \(\mathbf{r}'\): $$\text{area of } R = \int_{0}^{2\pi} \langle 0, -12 \sin{t}, 13 \cos{t} \rangle \cdot \langle 5 \cos{t}, -13 \sin{t}, 12 \cos{t} \rangle \, dt$$ Evaluate the dot product and integrate: $$\text{area of } R = \int_{0}^{2\pi} 0(5 \cos{t}) - (-12 \sin{t})(-13 \sin{t}) + (13 \cos{t})(12 \cos{t}) \, dt = \int_{0}^{2\pi} (-156 \sin^2{t} + 156 \cos^2{t}) \, dt$$ $$\text{area of } R = 156 \int_{0}^{2\pi} (\cos^2{t} - \sin^2{t}) \, dt$$ Using the double angle formula and changing the limits of integration, we obtain: $$\text{area of } R = 156 \int_{0}^{\pi} (1-2\sin^2{t})\, dt = 156 [t - \sin{(2t)}/2]_{0}^{\pi} = 156 \left(\pi - 0 \right) = 156\pi$$ Thus, the area of the region enclosed by the curve C is \(156\pi\).

Key concepts

Area of a region

The area of a region, denoted by \( R \), in a plane can often be determined using key mathematical principles such as Stokes' Theorem. This theorem is a powerful tool in vector calculus, which connects surface integrals of vector fields over a surface \( S \) to line integrals around the boundary curve \( C \) of that surface. In simple terms, Stokes' Theorem tells us that the area of \( R \) is equal to the line integral of the vector field \( \mathbf{F} \) around the curve \( C \).
To find the area using Stokes' Theorem, the formula is represented as:

\[ \oint_{C} \mathbf{F} \cdot d \mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dA \]
Here, \( abla \times \mathbf{F} \) represents the curl of \( \mathbf{F} \), and \( \mathbf{n} \) signifies the unit normal vector to the surface \( S \). Knowing that for a given region, the curl \( abla \times \mathbf{F} = \mathbf{n} \), simplifies our task, as the dot product \( \mathbf{n} \cdot \mathbf{n} = 1 \) thereby directly giving the area as \( \iint_{S} dA \), which is exactly the area of \( R \). Using such principles makes calculation of areas through vector fields precise and elegant.

Curl of a vector field

The curl of a vector field, represented by \( abla \times \mathbf{F} \), is a differential operator that describes the rotation or twisting of that field. To understand this in a simple way, think of the vector field as depicting fluid flow—where the curl at a point represents how much and in which direction the flow rotates around that point.
The mathematical computation of curl for a vector field \( \mathbf{F} = \langle F_x, F_y, F_z \rangle \) in three-dimensional space is given by:

\[ abla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \]
This yields a new vector that provides a thorough understanding of directional change. In the given exercise, computing the curl helped us establish it equal to the unit normal vector \( \mathbf{n} \), confirming that the vector field \( \mathbf{F} \) aligns with the plane region's orientation, thus making subsequent calculations, such as using Stokes' Theorem, effectively grounded.

Cross product

The cross product is an operation on two vectors in three-dimensional space that results in another vector that is perpendicular to the original two. The magnitude of the cross product corresponds to the area of the parallelogram that the vectors span. The direction follows the right-hand rule.
Mathematically, for vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the cross product is:

\[ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle \]

In the exercise, we used the cross product \( \mathbf{r} \times \mathbf{r'} \) when comparing derivatives of the position vector to test if the curve lies in a plane. The cross product being constant for all \( t \) means that the vector is always the same, implying that the curve \( C \) does not "jump" out of a plane. This fundamental property is exploited when working with geometrical shapes in three-dimensional space and helps in verifying planar uniformity.

Unit normal vector

A unit normal vector is a vector that is perpendicular to a surface at a given point and has a magnitude of 1. It plays a crucial role in vector calculus, particularly when applying Stokes' Theorem and dealing with surface integrals.
To find a unit normal vector for a given surface, you typically normalize another perpendicular vector. If you have a normal vector \( \mathbf{n} \), its unit version is calculated as:

\[ \frac{\mathbf{n}}{|\mathbf{n}|} \]
The dot product of any vector with itself yields its magnitude squared, but as the unit normal vector has a magnitude of 1, its dot product with itself remains 1, simplifying many calculations especially in integrals.
In the context of the exercise, finding the unit normal vector \( \mathbf{n} \) aligned with \( C \) was necessary to ensure that the direction of the curl of \( \mathbf{F} \) correctly relates to the surface defined by \( R \). This step fortified the foundation for evaluating the Stokes' Theorem integration correctly.