Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, use Stokes's Theorem to calculate \(\mathbf{F}=(z-y) \mathbf{i}+(z+x) \mathbf{j}-(x+y) \mathbf{k} ; S\) is the part of the paraboloid \(z=1-x^{2}-y^{2}\) above the \(x y\)-plane and \(\mathbf{n}\) is the upward normal.

Step-by-Step Solution

Answer

The integral is 0.

Step 1: Recall Stokes's Theorem

Stokes's Theorem states that \( \int_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \), where \(S\) is a surface bounded by curve \(C\), \(\mathbf{F}\) is a vector field, and \(abla \times \mathbf{F}\) is the curl of \(\mathbf{F}\). In this problem, \(S\) is the surface of the paraboloid \(z = 1 - x^2 - y^2\) above the xy-plane.

Step 2: Calculate the curl of \(\mathbf{F}\)

Find the curl of \(\mathbf{F} = (z-y) \mathbf{i} + (z+x) \mathbf{j} - (x+y) \mathbf{k}\). The curl is calculated as \(abla \times \mathbf{F} = \det\left(\begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \partial / \partial x & \partial / \partial y & \partial / \partial z \ z-y & z+x & -(x+y) \end{bmatrix}\right)\). Evaluate the determinant to find \(abla \times \mathbf{F} = (-1, -2, 1)\).

Step 3: Parametrize the boundary curve \(C\)

The boundary \(C\) of the surface \(S\) is a circle in the \(xy\)-plane where the paraboloid intersects the plane \(z = 0\). This occurs when \(1 - x^2 - y^2 = 0\), or \(x^2 + y^2 = 1\). Parametrize this circle using \(x = \cos(t)\) and \(y = \sin(t)\) for \(0 \leq t \leq 2\pi\).

Step 4: Evaluate the line integral

Substitute the parametrization of \(C\) into \(\mathbf{F}\): \(\mathbf{F}(\cos(t), \sin(t), 0) = (-\sin(t), \cos(t) + \cos(t), -(\cos(t) + \sin(t)))\). The differential \(d\mathbf{r}\) is \((-\sin(t), \cos(t), 0) dt\). Compute \(\oint_{C} \mathbf{F} \cdot d\mathbf{r}\) by evaluating the integral \(\int_{0}^{2\pi} (\mathbf{F} \cdot d\mathbf{r}) dt\). This evaluates to 0 after simplifying, which is consistent with \(abla \times \mathbf{F}\) being a constant and Stokes's Theorem.

Step 5: Conclude using Stokes's Theorem

According to Stokes's Theorem, since the line integral over the boundary is 0, the surface integral \(\int_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} = 0\) as well. Therefore, the result of the integral over the paraboloid surface is 0.

Key Concepts

Curl of a Vector Field

To understand how Stokes's Theorem applies, grasping the concept of a vector field's curl is vital. The curl of a vector field \( \mathbf{F} \) is a measure of its rotational motion at a point. It tells us how much the vector field 'twirls' around that point. In mathematical terms, the curl is represented by \( abla \times \mathbf{F} \).

In this exercise, our vector field is \( \mathbf{F}=(z-y) \mathbf{i}+(z+x) \mathbf{j}-(x+y) \mathbf{k} \). By using the determinant method, we find the curl as:

• Step 1: Set up the determinant with the \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) unit vectors, partial derivative operators \( \partial / \partial x \), \( \partial / \partial y \), and \( \partial / \partial z \), and the components of \( \mathbf{F} \).
• Step 2: Evaluate the determinant to get the curl \(abla \times \mathbf{F} = (-1, -2, 1) \).

Understanding the curl is crucial because Stokes's Theorem relates a line integral around a closed curve \( C \) to a surface integral of the curl of \( \mathbf{F} \) over the surface \( S \) bounded by \( C \).

Paraboloid Surface

The paraboloid surface is a beautiful shape often encountered in multivariable calculus. This surface is defined by the equation \( z = 1 - x^2 - y^2 \). It forms a dish-shaped curve that opens downward, lying above the \(xy\)-plane where \(z\) is positive.

In this problem, the surface \(S\) is that portion of the paraboloid that is above \(z = 0\). It's bounded by a circular edge, which is vital when applying Stokes's Theorem. This boundary circle is found where the paraboloid touches the \(xy\)-plane, or when \(z = 0\).

By setting \(1 - x^2 - y^2 = 0\), we identify this boundary as a circle with radius 1 centered at the origin: \(x^2 + y^2 = 1\). Mastering these surfaces helps in visualizing and connecting theoretical mathematics to physical shapes.

Line Integral

Line integrals allow us to integrate along a path or curve, rather than just over a traditional interval. This aspect is what connects the geometric features of Stokes's Theorem. In essence, a line integral calculates the work done by a vector field along a path.

For our problem, the path is the circle bounding the paraboloid, represented parametrically. We substitute the circle's parametric equations into the vector field, and then dot this vector with a differential arc length vector, \(d\mathbf{r}\).

• Parameters: \(x = \cos(t)\) and \(y = \sin(t)\) with \(0 \leq t \leq 2\pi\).
• Substitute into \(\mathbf{F}\) to get \((-\sin(t), 2\cos(t), -\cos(t) - \sin(t))\).
• Evaluate the integral over this path: \(\int_{0}^{2\pi} \mathbf{F} \cdot d\mathbf{r} dt\).

In this scenario, the integral ends up becoming zero, confirming that the net rotation around the curve is balanced, as per Stokes's Theorem.

Parametrization of Curves

Parametrization is a technique for expressing a curve using parameters, making them easier to handle mathematically. In our case, parametrization simplifies handling the boundary curve \(C\), the circle.

Consider the circle where the paraboloid meets the \(xy\)-plane. By parametrizing the curve, we transform \(x\) and \(y\) into functions of \(t\), where \(t\) typically varies from 0 to \(2\pi\).

• For a circle, you use \(x = \cos(t)\) and \(y = \sin(t)\).
• This approach maps out the entire circle as \(t\) loops from 0 around to \(2\pi\).

Parametrization is crucial in calculus because it simplifies the evaluation of line integrals, allowing for easier substitutions and integrations over complicated paths.