Question

Evaluate each integral in the simplest way possible. $$\iint \mathbf{P} \cdot \mathbf{n} d \sigma \text { over the upper half of the sphere } r=1 \text { if } \mathbf{P}=\operatorname{curl}(\mathbf{j} x-\mathbf{k} z)$$

Short answer

The integral evaluates to 0.

Step-by-step solution

- Identify the vector field \(\mathbf{P}\) as the curl

Given \(\mathbf{P} = \operatorname{curl}(\mathbf{j}x-\mathbf{k}z)\). \ Let's denote the vector field inside the curl as \(\mathbf{A} = \mathbf{j}x - \mathbf{k}z\). We need to find \(\operatorname{curl}(\mathbf{A})\).

- Calculate the curl of \(\mathbf{A}\)

The curl of a vector field \(\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}\) is given by: \[\operatorname{curl}(\mathbf{A}) = abla \times \mathbf{A} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right)\mathbf{i} + \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right)\mathbf{j} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right)\mathbf{k}\]For \(\mathbf{A} = \mathbf{j}x - \mathbf{k}z\), we have \(A_x = 0, A_y = x, A_z = -z\). Hence, \(\mathbf{P} = abla \times \mathbf{A} = \mathbf{i}(0 - 0) + \mathbf{j}(0 - 0) + \mathbf{k}(1 - 1) = \mathbf{0}\)

- Compute the surface integral

We need to evaluate the surface integral \(\iint \mathbf{P} \cdot \mathbf{n} d \sigma\). However, since \(\mathbf{P} = \mathbf{0}\), the integral simplifies to \[\iint \mathbf{0} \cdot \mathbf{n} d \sigma = 0\].

Key concepts

surface integrals

A surface integral allows us to integrate over a surface in three-dimensional space. It is used to find properties such as flux across a surface. Given a vector field \(\textbf{P}\), a surface integral over a surface \(\text{S}\) is written as: \[\iint_{S} \textbf{P} \cdot \textbf{n} \, d\sigma\] where \(\textbf{n}\) is the unit normal vector to the surface and \(d\sigma\) represents an infinitesimal element of area on the surface. The dot product \(\textbf{P} \cdot \textbf{n}\) finds the component of the vector field that is normal to the surface.
In our exercise, the goal was to evaluate the surface integral of the vector field \(\textbf{P}\) over the upper half of the sphere \(r = 1\). The result tells us how the vector field behaves across the surface.

curl of a vector field

The curl of a vector field measures the rotation or the twisting force at a point in the field. It is denoted as \(\operatorname{curl}(\textbf{A})\) or \(abla \times \textbf{A}\), where \(abla\) (nabla) is a vector differential operator. The curl in Cartesian coordinates for a vector field \(\textbf{A} = A_x \textbf{i} + A_y \textbf{j} + A_z \textbf{k} \) is given by: \[ abla \times \textbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ A_x & A_y & A_z \end{vmatrix} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \mathbf{k} \] In our exercise, for the vector field \(\textbf{A} = \textbf{j} x - \textbf{k} z\), we calculated the curl and found that it is zero everywhere, simplifying our integral significantly.

sphere in vector calculus

A sphere in vector calculus is often described using spherical coordinates. For a sphere with radius \(r\), the equation is \(x^2 + y^2 + z^2 = r^2\). Spherical coordinates are given by:

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

where \(\theta\) is the polar angle and \(\phi\) is the azimuthal angle.
For surface integrals over a sphere, it's crucial to consider the surface's orientation, represented by outward normal vectors. In our exercise, we dealt with the upper half of the sphere, simplifying calculations by focusing only on half the surface area.

Stokes' Theorem

Stokes' Theorem relates a surface integral over a surface \(S\) to a line integral over the boundary curve \(C\). It is written as: \[\iint_{S} \operatorname{curl}(\textbf{A}) \cdot \textbf{n} \, d\sigma = \oint_{C} \textbf{A} \cdot d\textbf{r}\] where:

• \(S\) is a smooth surface
• \(\textbf{n}\) is the unit normal vector to \(S\)
• \(C\) is the boundary curve of \(S\)
• \(d\textbf{r}\) is a vector tangent to \(C\)

In our exercise, Stokes' Theorem simplified evaluating the curl over a surface. By identifying that the curl of our vector field was zero, Stokes' Theorem confirmed the surface integral result as zero, reducing complexity greatly.