Let \( \alpha = \iint_S \vec{F} \cdot \hat{n} \, dS \), where \( \vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k} \) and \( S \) is the sphere with centre at \( (3, -1, 2) \) and radius \( 9 \). Here \( \hat{n} \) is the unit normal drawn outward and \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors.
Then the value of \( \dfrac{1}{36\pi}\alpha \) is equal to ______. (Answer in integer)
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Use the divergence theorem: for a closed surface, the flux equals the divergence integrated over the enclosed volume, and this divergence turns out to be constant here.
Step 1: Set up the divergence theorem.
$S$ is a closed sphere, so the flux integral converts into a volume integral by the divergence theorem: $\alpha = \iiint_V (\nabla \cdot \vec{F})\, dV$, where $V$ is the solid ball bounded by $S$.
Step 2: Compute the divergence.
Differentiate each component of $\vec{F} = (2x+3z)\hat{i}+(xz-y)\hat{j}+(y^2+2z)\hat{k}$ along its own axis and add:
\[ \nabla \cdot \vec{F} = 2 + (-1) + 2 = 3 \]
This is constant everywhere, so $\alpha = 3V$.
Step 3: Shift to the centre and switch to spherical coordinates.
Let $u=x-3, v=y+1, w=z-2$. This shift keeps volumes unchanged and turns the ball into $u^2+v^2+w^2 \le 81$. Using $u=r\sin\phi\cos\theta, v=r\sin\phi\sin\theta, w=r\cos\phi$, the volume element is $r^2\sin\phi\, dr\, d\phi\, d\theta$.
\[ \alpha = 3\int_0^{2\pi}\int_0^{\pi}\int_0^{9} r^2 \sin\phi \, dr\, d\phi\, d\theta \]
Step 4: Evaluate the three integrals one at a time.
\[ \int_0^9 r^2\, dr = \frac{729}{3} = 243, \qquad \int_0^{\pi}\sin\phi\, d\phi = 2, \qquad \int_0^{2\pi} d\theta = 2\pi \]
Multiply these with the constant $3$:
\[ \alpha = 3 \times 243 \times 2 \times 2\pi = 2916\pi \]
