We need to calculate the surface integral of the vector field F across the portion of the sphere $x^2 + y^2 + z^2 = a^2$ in the first octant, in the direction away from the origin.

Step 1: General formula for surface integral

The surface integral of a vector field $\mathbf{F}$ over a surface $S$ is given by: $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$ where $\mathbf{n}$ is the unit normal vector to the surface.

For the sphere, the outward unit normal vector is the normalized position vector: $\mathbf{n} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{(x, y, z)}{a}$

Step 2: Substitute $\mathbf{F}$ and $\mathbf{n}$

The vector field is: $\mathbf{F} = 7y \, \mathbf{i} - 7x \, \mathbf{j} + \mathbf{k}$ On the sphere $x^2 + y^2 + z^2 = a^2$, $|\mathbf{r}| = a$.

Dot product: $\mathbf{F} \cdot \mathbf{n} = \mathbf{F} \cdot \frac{\mathbf{r}}{a} = \frac{1}{a} \left( 7y x + (-7x y) + z \cdot 1 \right) = \frac{z}{a}$

Step 3: Convert to spherical coordinates

The sphere can be parameterized using spherical coordinates: $x = a \sin\phi \cos\theta, \quad y = a \sin\phi \sin\theta, \quad z = a \cos\phi$ where:

• $\phi$ is the polar angle ($0 \leq \phi \leq \frac{\pi}{2}$),
• $\theta$ is the azimuthal angle ($0 \leq \theta \leq \frac{\pi}{2}$).

In spherical coordinates, the surface element $dS$ is: $dS = a^2 \sin\phi \, d\phi \, d\theta$

Substituting $z = a \cos\phi$ into $\frac{z}{a}$, we have: $\mathbf{F} \cdot \mathbf{n} = \frac{a \cos\phi}{a} = \cos\phi$

The integral becomes: $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \int_0^{\pi/2} \int_0^{\pi/2} \cos\phi \, a^2 \sin\phi \, d\phi \, d\theta$

Step 4: Separate and evaluate the integral

Separate the integral into $\phi$ and $\theta$ components: $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = a^2 \int_0^{\pi/2} \int_0^{\pi/2} \cos\phi \sin\phi \, d\phi \, d\theta$

The $\phi$-integral is: $\int_0^{\pi/2} \cos\phi \sin\phi \, d\phi = \frac{1}{2} \int_0^{\pi/2} \sin(2\phi) \, d\phi = \frac{1}{2} \left[ -\frac{1}{2} \cos(2\phi) \right]_0^{\pi/2} = \frac{1}{2} \left( 0 - (-\frac{1}{2}) \right) = \frac{1}{4}$

The $\theta$-integral is: $\int_0^{\pi/2} d\theta = \frac{\pi}{2}$

Combine the results: $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = a^2 \cdot \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi a^2}{8}$

Final Answer:

$\boxed{\frac{\pi a^2}{8}}$

Would you like further clarification or a breakdown of any step?

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5 Related Questions:

1. How does the orientation of the normal vector affect the result of a surface integral?
2. What are the benefits of using spherical coordinates for surfaces like spheres?
3. Can you derive the formula for $dS$ in spherical coordinates?
4. How does this calculation change if the field or surface is in a different octant?
5. How would the calculation differ if the field direction was reversed?

Tip:

Always verify the domain of integration when working in spherical coordinates to avoid overcounting portions of the surface.