Question

If the outward flux of \( F(x, y, z) = (x^3, y^3, z^3) \) through the unit sphere \( x^2 + y^2 + z^2 = 1 \) is \( \alpha \pi \), then \( \alpha \) is equal to (round off to TWO decimal places):

Hint:

For flux integrals, apply the divergence theorem to simplify calculations using symmetry.

Answer 1

Step 1: Flux through the unit sphere. The flux is given by: \[ {Flux} = \int_{\partial V} F \cdot \hat{n} \, dS, \] where \( \hat{n} \) is the outward unit normal. Step 2: Divergence theorem. Using the divergence theorem: \[ {Flux} = \int_V (\nabla \cdot F) \, dV. \] Here, \( \nabla \cdot F = 3x^2 + 3y^2 + 3z^2 \). On the unit sphere, \( x^2 + y^2 + z^2 = 1 \), so \( \nabla \cdot F = 3 \). Step 3: Final computation. \[ {Flux} = \int_V 3 \, dV = 3 \cdot {Volume of the sphere} = 3 \cdot \frac{4\pi}{3} = 4\pi. \] Thus, \( \alpha = \frac{4}{\pi} \approx 2.41 \). Step 4: Conclusion. The value of \( \alpha \) is \( {2.41} \).