Water Flow Flux Calculation Explained

This problem asks us to calculate the total flow of water, also known as flux, through a specific rectangular area given the water flow rate vector.

Understanding the Water Flow Rate Vector

The water flow rate is given by the vector field $\vec{B} = 3yz \, \hat{i}$ liters/minute/m². This vector tells us the direction and magnitude of the flow at any point $(x, y, z)$ in space. The $\hat{i}$ component indicates that the flow is entirely in the positive x-direction. The magnitude of the flow rate depends on the y and z coordinates.

Identifying the Rectangular Area

The rectangular area is defined by the corners (0, 0, 0), (0, 3, 0), (0, 0, 2), and (0, 3, 2). Let's look at the coordinates:

• (0, 0, 0)
• (0, 3, 0)
• (0, 0, 2)
• (0, 3, 2)

Notice that the x-coordinate is 0 for all four points. This means the rectangular area lies in the yz-plane (where x=0). The vertices define a rectangle stretching from y=0 to y=3 and from z=0 to z=2 within the plane x=0.

The sides of the rectangle are along the y and z axes in the yz-plane. The side along the y-axis has length $3 - 0 = 3$ m. The side along the z-axis has length $2 - 0 = 2$ m. The area of this rectangle is $3 \times 2 = 6$ m².

Calculating the Total Flow (Flux)

The total flow or flux ($\Phi$) of a vector field $\vec{B}$ through a surface area A is calculated using the surface integral:

$$ \Phi = \iint_A \vec{B} \cdot \vec{dA} $$

Here, $\vec{B}$ is the water flow rate vector and $\vec{dA}$ is the differential area vector of the surface.

The rectangular area lies in the yz-plane (x=0). The flow is in the positive x-direction ($\hat{i}$). The area vector for a surface in the yz-plane that allows flow in the positive x-direction is in the positive x-direction. So, the differential area vector is $\vec{dA} = dA \, \hat{i}$. For a flat rectangular area in the yz-plane, $dA = dy \, dz$.

So, $\vec{dA} = dy \, dz \, \hat{i}$.

Now, let's calculate the dot product $\vec{B} \cdot \vec{dA}$:

$$ \vec{B} \cdot \vec{dA} = (3yz \, \hat{i}) \cdot (dy \, dz \, \hat{i}) $$

Since $\hat{i} \cdot \hat{i} = 1$, the dot product is:

$$ \vec{B} \cdot \vec{dA} = 3yz \, dy \, dz $$

The integration needs to be performed over the specified rectangular area, which spans from $y=0$ to $y=3$ and from $z=0$ to $z=2$.

The total flux is:

$$ \Phi = \int_{z=0}^{z=2} \int_{y=0}^{y=3} 3yz \, dy \, dz $$

Performing the Integration

We integrate with respect to y first:

$$ \int_{y=0}^{y=3} 3yz \, dy = 3z \int_{y=0}^{y=3} y \, dy $$

$$ = 3z \left[ \frac{y^2}{2} \right]_{y=0}^{y=3} $$

$$ = 3z \left( \frac{3^2}{2} - \frac{0^2}{2} \right) $$

$$ = 3z \left( \frac{9}{2} - 0 \right) $$

$$ = \frac{27}{2} z $$

Now, we integrate the result with respect to z from 0 to 2:

$$ \Phi = \int_{z=0}^{z=2} \frac{27}{2} z \, dz $$

$$ = \frac{27}{2} \int_{z=0}^{z=2} z \, dz $$

$$ = \frac{27}{2} \left[ \frac{z^2}{2} \right]_{z=0}^{z=2} $$

$$ = \frac{27}{2} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) $$

$$ = \frac{27}{2} \left( \frac{4}{2} - 0 \right) $$

$$ = \frac{27}{2} (2) $$

$$ = 27 $$

The total flow or flux of water through the rectangular area is 27 liters/minute.

Summary of Calculation Steps

1. Identify the water flow rate vector field $\vec{B}$.
2. Identify the surface area and its boundaries. Determine the differential area vector $\vec{dA}$.
3. Calculate the dot product $\vec{B} \cdot \vec{dA}$.
4. Set up the surface integral $\iint_A \vec{B} \cdot \vec{dA}$ with the correct limits of integration based on the area boundaries.
5. Evaluate the integral to find the total flow or flux.

Revision Table: Water Flow Flux

Additional Information: Surface Integrals and Flux

A surface integral is a generalization of multiple integrals to integration over surfaces. When the integrand is a vector field, the surface integral can represent the flux of the field through the surface.

• Flux: In physics and vector calculus, flux is a measure of the flow of a vector field through a surface. It is defined as the integral of the component of the vector field perpendicular to the surface over the area of the surface.
• Orientation of $\vec{dA}$: The direction of the differential area vector $\vec{dA}$ is normal (perpendicular) to the surface. For a closed surface, it typically points outwards. For an open surface like the rectangle here, the direction is usually specified or chosen based on the direction of the flow we are interested in. Since the flow $\vec{B}$ is in the $\hat{i}$ direction, and the surface is in the yz-plane (normal to $\hat{i}$), $\vec{dA}$ is parallel to $\vec{B}$ for positive flux in this case.
• Units of Flux: The units of flux are the units of the vector field multiplied by the units of area. In this problem, (liters/minute/m²) $\times$ (m²) gives liters/minute, which represents a volume flow rate.