Question

The motion of a liquid in a cylindrical container of radius 1 is described by the velocity field $$\mathbf{F}(x, y, z) .\( Find \)\int_{S} \int(\operatorname{curl} \mathbf{F}) \cdot \mathbf{N} d S$$ where \(S\) is the upper surface of the cylindrical container. \(\mathbf{F}(x, y, z)=-z \mathbf{i}+y \mathbf{k}\)

Short answer

The value of the surface integral is \(2\pi\).

Step-by-step solution

Calculate the curl of the vector field \(\mathbf{F}\)

The curl of a vector field \( \mathbf{F}=F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k} \) is given by \[ \operatorname{curl} \mathbf{F}= \left( \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z} \right) \mathbf{i}- \left( \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x} \right) \mathbf{j}+ \left( \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} \right) \mathbf{k} \]For \( \mathbf{F}=-z \mathbf{i}+y \mathbf{k} \), \( F_1=-z \), \( F_2=0 \), and \( F_3=y \). Thus,\[ \operatorname{curl} \mathbf{F}= \left( \frac{\partial y}{\partial y}-\frac{\partial 0}{\partial z} \right) \mathbf{i}- \left( \frac{\partial (-z)}{\partial z}-\frac{\partial y}{\partial x} \right) \mathbf{j}+ \left( \frac{\partial 0}{\partial x}-\frac{\partial (-z)}{\partial y} \right) \mathbf{k} = \mathbf{i} \]

Setting up the double integral

The curl of \( \mathbf{F} \) is a vector and the normal to the surface \( S \) is also a vector, so their dot product is a scalar. The surface integral of this scalar over the surface \( S \) is then a double integral in Cartesian coordinates. It can also be written in cylindrical coordinates by using that \( x=r \cos \theta \), \( y=r \sin \theta \) and \( dS=rdrd\theta \), where \( r \) is the radius of the cylinder and \( \theta \) is the angle from the positive x-axis. Thus, \[ \int_{S} (\operatorname{curl} \mathbf{F}) · \mathbf{N} dS = \int_{0}^{2\pi} \int_{0}^{1} (N\cdot \mathbf{i}) r dr d\theta \]The normal vector to the surface \( S \) is \( \mathbf{N}=\mathbf{i} \), so \( N\cdot \mathbf{i}=1 \). Thus,\[ \int_{S} (\operatorname{curl} \mathbf{F}) · \mathbf{N} dS = \int_{0}^{2\pi} \int_{0}^{1} r dr d\theta \]

Evaluating the double integral

Now, you can evaluate the remaining integral\[ \int_{0}^{2\pi} \int_{0}^{1} r dr d\theta = \int_{0}^{2\pi} \left[ \frac{1}{2}r^2 \right]_{0}^{1} d\theta = \int_{0}^{2\pi} \frac{1}{2} d\theta = \left[ \frac{1}{2}\theta \right]_{0}^{2\pi} = 2\pi - 0 =2\pi \]