Calculate the integral s f · ds, where s is the entire surface of the solid half ball x2 + y2 + z2 ≤ 1, z ≥ 0, and f = (x + 3y5)i + ( y + 10xz)j + (z − xy)k. (let s be oriented by the outward-pointing normal.)

Final answer:

To find the integral of the vector field over the solid half ball, we calculate two surface integrals, one over the flat disk and one over the hemisphere's curved surface, and then sum their contributions.

Explanation:

The student has asked to calculate the surface integral of the vector field f = (x + 3y^5)i + ( y + 10xz)j + (z - xy)k over the upper half of the sphere [tex]x^2 + y^2 + z^2 = 1[/tex], with z ≥ 0. To solve this, we can use the divergence theorem, noting that the divergence of f is ∇ · f, and then integrating that over the volume of the half sphere. We divide the surface S into two parts: the flat circular disk at the bottom (where z = 0) and the curved surface of the half sphere. The flux through the curved surface can be found by integrating over the sphere's surface, while the disk contributes a separate integral. This requires calculating two different surface integrals, and adding their results.

The surface integral [tex]\(\int_{S} \mathbf{f} \cdot d\mathbf{s}\)[/tex] for the given vector field and half-ball surface is [tex]\(2\pi\)[/tex].

To calculate the surface integral [tex]\(\int_{S} \mathbf{f} \cdot d\mathbf{s}\)[/tex] where [tex]\(S\)[/tex] is the entire surface of the solid half-ball [tex]\(x^2 + y^2 + z^2 \leq 1, z \geq 0\) and \(\mathbf{f} = (x + 3y^5)\mathbf{i} + (y + 10xz)\mathbf{j} + (z - xy)\mathbf{k}\)[/tex], we will use the Divergence Theorem.

The Divergence Theorem states that for a vector field [tex]\(\mathbf{f}\)[/tex],

[tex]\[\int_{S} \mathbf{f} \cdot d\mathbf{s} = \int_{V} (\nabla \cdot \mathbf{f}) \, dV\][/tex]

where [tex]\(S\)[/tex] is the boundary surface of the volume [tex]\(V\),[/tex] oriented by the outward-pointing normal.

1. Calculate the Divergence [tex]\(\nabla \cdot \mathbf{f}\)[/tex]:

[tex]\[\nabla \cdot \mathbf{f} = \frac{\partial}{\partial x}(x + 3y^5) + \frac{\partial}{\partial y}(y + 10xz) + \frac{\partial}{\partial z}(z - xy)\][/tex]

Computing each term:

[tex]\[\frac{\partial}{\partial x}(x + 3y^5) = 1\][/tex]

[tex]\[\frac{\partial}{\partial y}(y + 10xz) = 1\][/tex]

[tex]\[\frac{\partial}{\partial z}(z - xy) = 1\][/tex]

So the divergence is:

[tex]\[\nabla \cdot \mathbf{f} = 1 + 1 + 1 = 3\][/tex]

2. Set up the volume integral:

The volume (V) is the solid half-ball defined by [tex]\(x^2 + y^2 + z^2 \leq 1\)[/tex] and [tex]\(z \geq 0\)[/tex].

We will integrate the constant divergence \(3\) over the volume of the half-ball. The volume of a full ball of radius 1 is \(\frac{4}{3} \pi (1^3) = \frac{4}{3} \pi\). Since we have a half-ball, the volume is:

[tex]\[\text{Volume of half-ball} = \frac{1}{2} \cdot \frac{4}{3} \pi = \frac{2}{3} \pi\][/tex]

3. Evaluate the volume integral:

[tex]\[\int_{V} (\nabla \cdot \mathbf{f}) \, dV = \int_{V} 3 \, dV = 3 \int_{V} dV = 3 \cdot \text{Volume of half-ball} = 3 \cdot \frac{2}{3} \pi = 2 \pi\][/tex]

Therefore, the surface integral is:

[tex]\[\int_{S} \mathbf{f} \cdot d\mathbf{s} = 2\pi\][/tex]

So, the final answer is:

[tex]\[\boxed{2\pi}\][/tex]

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