Question

In Exercises \(13-16,\) a closed surface \(S\) and a vector field \(\vec{F}\) are given. Find the outward flux of \(\vec{F}\) over \(S\) either through direct computation or through the Divergence Theorem. $$ \begin{aligned} &\mathcal{S} \text { is the surface formed by the intersections of the planes }\\\ &z=\frac{1}{2}(3-x), x=1, y=0, y=2 \text { and } z=0 ; \vec{F}=\left\langle x, y^{2}, z\right\rangle \end{aligned} $$

Short answer

The outward flux of \( \vec{F} \) over \( S \) is 10.

Step-by-step solution

Understand the Surface S

The surface \( S \) is described by the intersection of several planes: \( z = \frac{1}{2}(3-x) \), \( x = 1 \), \( y = 0 \), \( y = 2 \), and \( z = 0 \). These equations define a region bounded in three-dimensional space. This region is a prism with a sloped top surface defined by \( z = \frac{1}{2}(3-x) \), and the vertical sides are defined by the planes \( x = 1 \), \( y = 0 \), and \( y = 2 \). The base of the prism is defined by the plane \( z = 0 \).

Check if Divergence Theorem is Applicable

The Divergence Theorem can be applied if the vector field \( \vec{F} \) and the surface \( S \) satisfy smoothness and closedness conditions. The surface \( S \) is closed, as it forms a shape encapsulating a volume in three dimensions. The vector field \( \vec{F} = \langle x, y^2, z \rangle \) is continuous and smooth in this region, making the theorem applicable.

Compute the Divergence of F

The Divergence Theorem relates surface integrals over a closed surface \( S \) to volume integrals over the volume \( V \) enclosed by \( S \). First, find the divergence of \( \vec{F} \):\[ abla \cdot \vec{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z) = 1 + 2y + 1 = 2y + 2. \]

Set Up and Evaluate the Volume Integral

The volume \( V \) is enclosed by limits: \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 2 \), and for a given \( x \), \( z \) ranges from \( 0 \) at the base to \( \frac{1}{2}(3-x) \) at the top plane. Set up the triple integral:\[ \int_V (2y + 2) \, dV = \int_{x=0}^{1} \int_{y=0}^{2} \int_{z=0}^{\frac{1}{2}(3-x)} (2y + 2) \, dz \, dy \, dx. \]

Integrate With Respect to z

Integrate the expression with respect to \( z \):\[ \int_{z=0}^{\frac{1}{2}(3-x)} (2y + 2) \, dz = (2y + 2) \left[ z \right]_{0}^{\frac{1}{2}(3-x)} = (2y + 2) \left( \frac{1}{2}(3-x) \right). \]

Integrate With Respect to y

Now integrate with respect to \( y \):\[ \int_{y=0}^{2} (2y+2)\left(\frac{1}{2}(3-x)\right) \, dy = \left(\frac{1}{2}(3-x)\right) \int_{y=0}^{2} (2y+2) \, dy. \]Calculate the inner integral:\[ \int_{y=0}^{2} (2y+2) \, dy = [y^2 + 2y]_{0}^{2} = (4+4) - (0+0) = 8. \]Thus, the expression becomes \( 8\left(\frac{1}{2}(3-x)\right) \).

Integrate With Respect to x

Integrate the remaining expression with respect to \( x \):\[ \int_{x=0}^{1} 8 \left(\frac{1}{2}(3-x)\right) \, dx = 4 \left[ 3x - \frac{x^2}{2} \right]_{0}^{1} = 4 \left(3 \cdot 1 - \frac{1^2}{2} - 0 \right) = 4 \left( 3 - \frac{1}{2} \right) = 4 \cdot \frac{5}{2} = 10. \]

Conclusion

The value of the outward flux of \( \vec{F} \) over the surface \( S \) is 10, using the Divergence Theorem.

Key concepts

Vector Field

In calculus and physics, a vector field assigns a vector to each point in a subset of space. Imagine each point in a region having a tiny arrow attached to it. The direction and length of the arrow represent different fields, such as fluid flows, electromagnetic fields, or in our case, a mathematical vector field. The exciting aspect is how these vectors behave and interact with a surface in three-dimensional space. In this exercise, we considered the vector field \( \vec{F} = \langle x, y^2, z \rangle \). Each component of this vector field depends on one spatial variable: \( x \), \( y \), and \( z \).

• The first component, \( x \), indicates the vector field has uniform strength along the \( x \)-direction.
• The second component, \( y^2 \), describes growth along the \( y \)-axis.
• The third is \( z \), adding direction pointing upwards.

The combination of these components determines the net flow or influence exerted around a surface. Understanding this basic structure helps in performing calculations, such as finding the outward flux through a surface.

Outward Flux

Outward flux is a concept that tells us how much of the vector field passes through a closed surface in an outward direction. It can be thought of like a net flow or the total amount of a field flowing out of the surface. Imagine standing inside a bubble and watching how the field emerges out of it. This is quantified using mathematical operations.

To find the outward flux, either compute directly through surface integrals or utilize the Divergence Theorem, which relates outward flux to a volume integral over the enclosed region. Here, we use the Divergence Theorem which simplifies these calculations for closed surfaces. The key calculation involves finding the divergence \( abla \cdot \vec{F} \). For our vector field \( \vec{F} = \langle x, y^2, z \rangle \), this gives a result of \( 2y + 2 \), representing how much the field expands in three-dimensional space.

Surface Integrals

Surface integrals are a mathematical way of generalizing functions over a surface. They are like adding up all the infinitesimal products of the field and the surface area, taking into account the direction. This is especially useful in physics and engineering to calculate things like total force or flux.

Calculating surface integrals directly can be complex, as it involves parametrizing the surface and integrating the field across this parametric description. However, by using the Divergence Theorem, we transform the surface integral into a volume integral, reducing complexity. The theorem works well here since the surface \( S \) encloses a volume, and the field \( \vec{F} \) behaves smoothly within this region. Thus, we efficiently solve for the outward flux via volume integration instead.

Volume Integrals

Volume integrals extend the idea of integration to three dimensions. Instead of adding up values along a line or across a surface, we sum them through a volume. This makes them vital for applying the Divergence Theorem.

When working with volume integrals, as in this exercise, you assign limits corresponding to the region's dimensions in space. For the prism described in the problem, we consider limits for \( x \), \( y \), and \( z \), reflecting how \( z \) changes from its base to the top plane. Integrating the divergence \( 2y+2 \) over the described volume, you compute the total spread of the field within, resulting in the outward flux through the enclosing surface.
Volume integrals offer a comprehensive way to sum up every part of the field within the bounded space, making it feasible to calculate phenomena like fluid flow rates or electromagnetic flux.