Problem 354 - Calculus Volume 3

Question

Use the surface integral in Stokes' theorem to \(\begin{array}{llll}\text { calculate } & \text { the } & \text { circulation } & \text { of } & \text { field }\end{array}\) \(\mathbf{F}(x, y, z)=x^{2} y^{3} \mathbf{i}+\mathbf{j}+z \mathbf{k}\) around \(C,\) which is the intersection of cylinder \(x^{2}+y^{2}=4\) and hemisphere \(x^{2}+y^{2}+z^{2}=16, z \geq 0, \quad\) oriented counterclockwise when viewed from above.

Step-by-Step Solution

Answer

The circulation of \( \mathbf{F} \) around curve \( C \) is zero.

Step 1: Understand Stokes' Theorem

Stokes' Theorem relates a surface integral of a curl of a vector field to a line integral over the surface's boundary. Mathematically, it is expressed as \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} \), where \( S \) is the surface bounded by the curve \( C \). In this exercise, \( C \) is the intersection of a cylinder and a hemisphere.

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

The vector field is given by \( \mathbf{F}(x, y, z) = x^2 y^3 \mathbf{i} + \mathbf{j} + z \mathbf{k} \). To apply Stokes' theorem, find the curl \( abla \times \mathbf{F} \).Use the determinant form:\[abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \x^2 y^3 & 1 & z \end{vmatrix}\]This gives:\[abla \times \mathbf{F} = (3x^2 y^2) \mathbf{i} - (2xy^3) \mathbf{k}\]

Step 3: Define the Surface \( S \)

Choose the hemisphere as the surface \( S \), since it is easily parameterized and bounded by the intersection curve \( C \). The surface is given by \( x^2 + y^2 + z^2 = 16 \) with \( z \geq 0 \). Parametrize the hemisphere using spherical coordinates: \( x = 4 \sin \theta \cos \phi \), \( y = 4 \sin \theta \sin \phi \), \( z = 4 \cos \theta \), where \( 0 \leq \theta \leq \frac{\pi}{2} \) and \( 0 \leq \phi < 2\pi \).

Step 4: Calculate the Differential Surface Area \( d\mathbf{S} \)

The differential surface area in spherical coordinates is \( d\mathbf{S} = 4 \sin \theta \, d\theta \, d\phi \cdot \mathbf{n} \), where \( \mathbf{n} \) is the outward normal to the surface. Compute the normal using the gradient of \( f(x, y, z) = x^2 + y^2 + z^2 - 16 \):\[ abla f = 2x \mathbf{i} + 2y \mathbf{j} + 2z \mathbf{k} \]At \( (x, y, z) \), the normal \( \mathbf{n} = (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) / 4 \). Since we're considering upward orientation, use the positive \( \mathbf{k} \) direction.

Step 5: Compute the Surface Integral

Substitute \( abla \times \mathbf{F} \) and \( d\mathbf{S} \) into the surface integral:\[\iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{S} (3x^2 y^2 \mathbf{i} - 2xy^3 \mathbf{k}) \cdot (0\mathbf{i} + 0\mathbf{j} + 4\sin\theta \, \hat{\mathbf{k}}) \, \sin\theta \, d\theta \, d\phi\]Calculate the dot product:\[ \iint_{S} -2xy^3 \cdot 4\sin\theta \, d\theta \, d\phi = -8 \iint_{S} xy^3 \sin\theta \, d\theta \, d\phi \]This integral vanishes because the integrand \( xy^3 \sin\theta \) is odd with respect to the symmetrical limits of azimuthal angle \( \phi \).

Step 6: Solution Conclusion

Since the surface integral evaluates to zero due to the odd nature of the integrand, by Stokes' theorem, this implies the circulation of \( \mathbf{F} \) around curve \( C \) is also zero.

Key Concepts

Surface Integral

To understand the concept of a surface integral, imagine trying to calculate the total of a quantity over a curved surface. Instead of summing up over a flat area like in typical integrals, you consider the entire 3D surface.
A surface integral can measure various physical quantities such as the flow through a surface. It is used in Stokes' Theorem to connect a vector field's behavior over a surface to a curve boundary.

• In Stokes' Theorem, the surface integral of the curl of a vector field over a surface is equal to the line integral of the vector field around the boundary of the surface.
• This makes it a powerful tool for converting between two different types of integrals.
• In the exercise, we used the surface integral to determine the circulation around the curve.

Curl of a Vector Field

The curl of a vector field is a measure of how the field "twists" or "rotates" around a point. It essentially describes the field's tendency to rotate around a given point.
To calculate the curl, we use a determinant form involving partial derivatives of the vector field's components. In mathematical terms, it is given by:
\[abla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k}\].
For our vector field \( \mathbf{F}(x, y, z) = x^2 y^3 \mathbf{i} + \mathbf{j} + z \mathbf{k} \), the curl is calculated as:
\[abla \times \mathbf{F} = (3x^2 y^2) \mathbf{i} - (2xy^3) \mathbf{k}\].

Parametrization of Surfaces

Parametrization is a way to describe a surface using a set of equations that map a region in a parameter space to points on the surface. This approach helps in working with complex curved shapes like spheres, cylinders, or hemispheres.
In this exercise, the hemisphere's surface, defined by \(x^2 + y^2 + z^2 = 16\) with \(z \geq 0\), is parametrized using spherical coordinates:

• \(x = 4 \sin \theta \cos \phi\)
• \(y = 4 \sin \theta \sin \phi\)
• \(z = 4 \cos \theta\)

Where \(0 \leq \theta \leq \frac{\pi}{2}\) and \(0 \leq \phi < 2\pi\).
This parametrization covers the entire upper hemisphere and allows us to express complex surface integrals more conveniently.

Differential Surface Area

Differential surface area \(d\mathbf{S}\) determines a tiny patch on a parameterized surface that contributes to the entire surface integral operation. It is represented by vectors that point in a direction perpendicular to the surface, reflecting the orientation.
In spherical coordinates for our hemisphere, the differential surface area is represented as:
\[d\mathbf{S} = 4 \sin \theta \, d\theta \, d\phi \cdot \mathbf{n}\], where \(\mathbf{n}\) is the normal vector of the surface, and is obtained from the gradient of the surface function \(f(x, y, z) = x^2 + y^2 + z^2 - 16\):
\[abla f = 2x \mathbf{i} + 2y \mathbf{j} + 2z \mathbf{k} \].
By calculating at \((x, y, z)\), the normal \(\mathbf{n} = (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) / 4\). The normal direction matches the orientation we're using for Stokes' Theorem calculation.
Proper understanding of differential surface area ensures accurate integration over the surface.