Problem 346 - Calculus Volume 3

Question

346\. Use Stokes' theorem to evaluate \(\iint_{S}(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}) d S,\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}+y^{2} \mathbf{j}+z e^{x y} \mathbf{k}\) and \(S\) is the part of surface \(z=1-x^{2}-2 y^{2}\) with \(z \geq 0\), oriented counterclockwise.

Step-by-Step Solution

Answer

The value of the integral is \(\frac{\pi^2}{4\sqrt{2}}\).

Step 1: Understand Stokes' Theorem

Stokes' Theorem states that \( \iint_{S} (\text{curl } \mathbf{F} \cdot \mathbf{N}) \, dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \), where \(S\) is a surface bounded by a curve \(C\) and \(\mathbf{N}\) is the unit normal to the surface. We will use this to evaluate the given surface integral over \(S\).

Step 2: Find the Curl of \(\mathbf{F}\)

The vector field given is \( \mathbf{F}(x, y, z) = x \mathbf{i} + y^2 \mathbf{j} + ze^{xy} \mathbf{k} \). The curl of \(\mathbf{F}\), \( \text{curl } \mathbf{F} = abla \times \mathbf{F} \), is calculated as:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \x & y^2 & ze^{xy}\end{vmatrix}\]Calculating the determinant, we find:\( \text{curl } \mathbf{F} = (y^2 e^{xy} - ze^{xy} y) \mathbf{i} - (e^{xy} - ze^{xy} x) \mathbf{j} + (2y - 0) \mathbf{k} = (y^2 e^{xy} - yze^{xy}) \mathbf{i} + (ze^{xy} x - e^{xy}) \mathbf{j} + 2y \mathbf{k} \).

Step 3: Describe the Surface \(S\)

The surface \(S\) is the part of the paraboloid \(z = 1 - x^2 - 2y^2\) where \(z \geq 0\). In the \(xy\)-coordinate plane, this corresponds to the region \(R\) given by \( x^2 + 2y^2 \leq 1 \). The boundary curve \(C\) is the ellipse where \(z = 0\), i.e., \( x^2 + 2y^2 = 1 \).

Step 4: Parameterize the Boundary Curve \(C\)

The boundary \(C\) of the surface can be parameterized as an ellipse:\[ \mathbf{r}(t) = (\cos t) \mathbf{i} + (\frac{1}{\sqrt{2}} \sin t) \mathbf{j}, \quad t \in [0, 2\pi] \].Calculate \( d\mathbf{r} = (\frac{d}{dt} (\cos t)) \mathbf{i} + (\frac{d}{dt} \frac{1}{\sqrt{2}} \sin t) \mathbf{j} = -\sin t \mathbf{i} + \frac{1}{\sqrt{2}} \cos t \mathbf{j} \, dt \).

Step 5: Evaluate the Line Integral

Using Stokes' Theorem, we calculate \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \).Substitute \( \mathbf{F} \) and \( d\mathbf{r} \): \( \oint_{C} (x \mathbf{i} + y^2 \mathbf{j} + ze^{xy} \mathbf{k}) \cdot (-\sin t \mathbf{i} + \frac{1}{\sqrt{2}} \cos t \mathbf{j}) \, dt \).Evaluate this integral using the parameterization:\[ \oint_{0}^{2\pi} \left( (-\cos t \sin t) + \frac{1}{\sqrt{2}} (\frac{1}{2} \sin^2 t) \right) \, dt \].This simplifies to the integral over one period of trigonometric terms, which can typically be reduced using standard integral results.

Step 6: Simplify and Finalize the Integral

Integrating the expression:\( \oint_{0}^{2\pi} \left( (-\cos t \sin t) + \frac{1}{4\sqrt{2}} \sin^2 t \right) \, dt \), each term has standard integral solutions over the period \([0, 2\pi]\):- The integral of \(-\cos t \sin t\) is zero over \([0, 2\pi]\).- The integral of \(\sin^2 t\) can be solved using the identity \(\sin^2 t = \frac{1 - \cos(2t)}{2}\), leading to a final value of \(\pi\) from \([0, 2\pi]\).Thus, the integral evaluates to \(\frac{\pi}{4\sqrt{2}} \cdot \pi = \frac{\pi^2}{4\sqrt{2}}\).

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on them. It is essential in physics and engineering, especially where multiple quantities and directions are involved. Vector operations include differentiation and integration over various dimensions.

Key operations in vector calculus are:

• Gradient: Gives the rate and direction of change of a scalar field.
• Divergence: Measures how much a vector field "spreads out" from a point.
• Curl: Describes the rotation of a vector field at a point.

Understanding these operations is vital, as they help describe physical phenomena such as fluid flow, electromagnetic fields, and more. Stokes' Theorem, which involves the curl, is a critical application of vector calculus in computing integrals over surfaces.

Surface Integrals

Surface integrals extend the concept of integrals to functions over surfaces. They calculate the total effect (such as flux or mass) over a surface in three-dimensional space. When dealing with vector fields, surface integrals become particularly powerful.

To perform a surface integral, follow these key steps:

• Parameterize the surface with variables that trace over it.
• Compute the vector differential area element, which involves the normal vector.
• Evaluate the integral directly or use theorems like Stokes' or Gauss's Theorem for simplification.

Using surface integrals, it becomes possible to analyze complex surfaces, which is crucial for fields like fluid mechanics and electromagnetism.

Curl of a Vector Field

The curl of a vector field provides a measure of rotation or swirling at each point in the field. It is crucial when understanding rotational fields in physics, such as fluid vortices or magnetic fields.

The curl \( abla \times \mathbf{F} \) of a vector field \( \mathbf{F}(x, y, z) = P(x, y, z)\mathbf{i} + Q(x, y, z)\mathbf{j} + R(x, y, z)\mathbf{k} \) is:\[(\partial_y R - \partial_z Q) \mathbf{i} + (\partial_z P - \partial_x R) \mathbf{j} + (\partial_x Q - \partial_y P) \mathbf{k}\]The resulting vector field indicates direction and magnitude of rotation, useful in fluid dynamics and electromagnetism. Calculating the curl simplifies many surface integrals via Stokes' Theorem, as it allows conversion to line integrals along a boundary.

Line Integrals

Line integrals compute the value of a function along a curve, summing up effects over one-dimensional paths. They're heavily employed in physics, where forces, electricity, and fluid flow need evaluation over specific paths.

Calculating line integrals involves:

• Parameterizing the path with a variable, typically \( t \), tracing the curve.
• Expressing the vector field along the curve and adding the contributions at each point.
• Integrating over the path between desired limits.

Line integrals also connect to Stokes' Theorem, enabling the conversion of complicated surface integrations into simpler one-dimensional path evaluations. Notably, they help compute work done by forces along a path, central to classical mechanics and electromagnetism.