Question

Consider a sphere of radius 4, centered at the origin, with outward unit normal $\hat{n}$ on its surface $S$. The value of the surface integral $\iint_{S}\dfrac{2x\hat{i}+3y\hat{j}+4z\hat{k}}{4\pi}\cdot \hat{n}\, dA$ is \(\underline{\hspace{2cm}}\) (rounded off to one decimal place).

Hint:

For symmetric surfaces like spheres, use known integral identities or symmetry to simplify surface integrals.

Answer 1

Correct Answer: 191.9 - 192.1

Given vector field:
$\vec{F} = \dfrac{1}{4\pi}(2x\hat{i} + 3y\hat{j} + 4z\hat{k})$
For a sphere centered at origin:
$\hat{n} = \dfrac{\vec{r}}{|\vec{r}|} = \dfrac{x\hat{i} + y\hat{j} + z\hat{k}}{4}$
Dot product:
$\vec{F} \cdot \hat{n} = \dfrac{1}{4\pi}(2x, 3y, 4z) \cdot \dfrac{1}{4}(x,y,z)$
$= \dfrac{1}{16\pi}(2x^{2} + 3y^{2} + 4z^{2})$
On radius 4 sphere:
$x^{2} + y^{2} + z^{2} = 16$
Average value weighted by coefficients:
$2x^{2} + 3y^{2} + 4z^{2}$ integrated over sphere gives:
$\iint_{S} (2x^{2} + 3y^{2} + 4z^{2})\, dA = (2 + 3 + 4)\dfrac{4\pi r^{4}}{3}$
Compute:
$= 9 \cdot \dfrac{4\pi (4^{4})}{3}$
$= 9 \cdot \dfrac{4\pi \cdot 256}{3}$
$= 9 \cdot \dfrac{1024\pi}{3}$
$= 3072\pi$
Now multiply by prefactor:
$\iint_{S} \vec{F} \cdot \hat{n}\, dA = \dfrac{1}{16\pi} \cdot 3072\pi$
$= \dfrac{3072}{16} = 192$
Rounded to one decimal place: $192.0$