Flux Through A Semi-Sphere: A Vector Field Problem

Hey guys! Today, we're diving into a super interesting problem involving vector fields and flux calculation across a semi-sphere. This is a classic example that pops up in vector calculus, and understanding it will seriously level up your problem-solving skills. We'll break it down step by step, so don't worry if it seems intimidating at first. Let's get started!

Understanding the Problem

So, the problem statement is: Let $S$ be the semi-sphere with equation $z = \sqrt{1-x^2-y^2}$, oriented by the normal vector whose $z$ component is positive. Determine the flux through $S$ of the vector field $\mathbf{F}(x, y, z) = (e^{y^2} \cos(z) + x, e^{x^2} \sin(z) + y, z+1)$.

What does all of that mean? Let's dissect it. The semi-sphere $S$ is the top half of a sphere centered at the origin with radius 1. The orientation given by the normal vector with a positive $z$ component just means we're considering the outward direction. We're asked to find the flux of the vector field $\mathbf{F}$ through this surface. In simpler terms, we want to know how much of the vector field is "flowing" through the semi-sphere.

Key Concepts

Before we dive into the calculations, let's quickly recap some essential concepts:

• Vector Field: A vector field assigns a vector to each point in space. Think of it as representing a force or velocity at every location.
• Flux: The flux of a vector field through a surface measures the amount of the field passing through the surface. Mathematically, it's the surface integral of the vector field's normal component.
• Divergence Theorem: This is a big deal for this problem. It relates the flux of a vector field through a closed surface to the volume integral of the divergence of the field over the region enclosed by the surface. In simpler terms, it turns a surface integral into a volume integral, which can be much easier to compute.

Setting up the Solution

The most efficient way to solve this problem is by using the Divergence Theorem. This theorem allows us to convert the surface integral (which is usually a pain to compute directly) into a volume integral. The Divergence Theorem states:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV$

Where:

• $S$ is a closed surface.
• $V$ is the volume enclosed by $S$.
• $\mathbf{F}$ is a vector field.
• $\nabla \cdot \mathbf{F}$ is the divergence of $\mathbf{F}$.

Applying the Divergence Theorem

To apply the Divergence Theorem, we need a closed surface. Our semi-sphere $S$ isn't closed; it's open at the bottom. So, we'll close it by adding a disk $D$ in the $xy$-plane, defined by $x^2 + y^2 \le 1$ and $z = 0$. Let's call the closed surface $S \cup D$.

Now, we can apply the Divergence Theorem to the closed surface $S \cup D$:

$\iint_{S \cup D} \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV$

This means:

$\iint_S \mathbf{F} \cdot d\mathbf{S} + \iint_D \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV$

Our goal is to find $\iint_S \mathbf{F} \cdot d\mathbf{S}$, so we can rearrange the equation:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV - \iint_D \mathbf{F} \cdot d\mathbf{S}$

Calculating the Divergence

The first step is to calculate the divergence of the vector field $\mathbf{F}(x, y, z) = (e^{y^2} \cos(z) + x, e^{x^2} \sin(z) + y, z+1)$. The divergence is given by:

$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(e^{y^2} \cos(z) + x) + \frac{\partial}{\partial y}(e^{x^2} \sin(z) + y) + \frac{\partial}{\partial z}(z+1)$

Calculating the partial derivatives:

• $\frac{\partial}{\partial x}(e^{y^2} \cos(z) + x) = 1$
• $\frac{\partial}{\partial y}(e^{x^2} \sin(z) + y) = 1$
• $\frac{\partial}{\partial z}(z+1) = 1$

Therefore, the divergence is:

$\nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3$

Evaluating the Volume Integral

Now we need to evaluate the volume integral $\iiint_V (\nabla \cdot \mathbf{F}) dV$. Since $\nabla \cdot \mathbf{F} = 3$, we have:

$\iiint_V 3 dV = 3 \iiint_V dV$

The integral $\iiint_V dV$ represents the volume of the region $V$, which is the volume of the half-sphere with radius 1. The volume of a full sphere is $\frac{4}{3}\pi r^3$, so the volume of the half-sphere is:

$V = \frac{1}{2} \cdot \frac{4}{3}\pi (1)^3 = \frac{2}{3}\pi$

Thus, the volume integral is:

$3 \iiint_V dV = 3 \cdot \frac{2}{3}\pi = 2\pi$

Evaluating the Surface Integral over the Disk

Next, we need to evaluate the surface integral $\iint_D \mathbf{F} \cdot d\mathbf{S}$ over the disk $D$. The disk lies in the $xy$-plane, so its normal vector points in the negative $z$ direction (since we're closing the surface outward). Thus, $d\mathbf{S} = (0, 0, -1) dA$.

On the disk $D$, $z = 0$, so the vector field $\mathbf{F}$ becomes:

$\mathbf{F}(x, y, 0) = (e^{y^2} \cos(0) + x, e^{x^2} \sin(0) + y, 0+1) = (e^{y^2} + x, y, 1)$

Now we compute the dot product $\mathbf{F} \cdot d\mathbf{S}$:

$\mathbf{F} \cdot d\mathbf{S} = (e^{y^2} + x, y, 1) \cdot (0, 0, -1) = -1$

So, the surface integral over the disk is:

$\iint_D \mathbf{F} \cdot d\mathbf{S} = \iint_D -1 dA = - \iint_D dA$

The integral $\iint_D dA$ represents the area of the disk $D$, which is a circle with radius 1. The area of a circle is $\pi r^2$, so the area of the disk is:

$A = \pi (1)^2 = \pi$

Thus, the surface integral is:

$- \iint_D dA = -\pi$

Finding the Flux Through the Semi-Sphere

Finally, we can plug everything back into our equation:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV - \iint_D \mathbf{F} \cdot d\mathbf{S}$

$\iint_S \mathbf{F} \cdot d\mathbf{S} = 2\pi - (-\pi) = 2\pi + \pi = 3\pi$

Therefore, the flux of the vector field $\mathbf{F}$ through the semi-sphere $S$ is $3\pi$.

Conclusion

Woohoo! We made it! By applying the Divergence Theorem, we transformed a seemingly complex surface integral problem into a much more manageable volume integral and a surface integral over a disk. Remember, the key is to close the surface, calculate the divergence, and carefully evaluate the resulting integrals. I hope this breakdown helped you understand how to tackle these kinds of problems. Keep practicing, and you'll become a vector calculus whiz in no time! Keep rocking!