Question

Evaluate ∫∫ A .nˆ dS , where A = x (cos)^2 y ˆi + xz ˆj + z (sin)^2y kˆ , over the surface of a sphere
with its centre at the origin and radius of 4 units.
note ^i=icap(unit vector) so on for other

Answer 1

Answer on Question#39178 Math - Other

Evaluate $\iint \vec{A} \vec{n} dS$ , where $\vec{A} = x\cos^2 y\vec{i} + xz\vec{j} + z\sin^2 y\vec{k}$ , over the surface of a sphere with its centre at the origin and radius of 4 units.

Solution:

$$\iint \vec{A} \vec{n} \, dS = \iiint \operatorname{div} \vec{A} \, dx \, dy \, dz$$ $$\operatorname{div} \vec{A} = \frac{\partial (x \cos^2 y)}{\partial x} + \frac{\partial (xz)}{\partial y} + \frac{\partial (z \sin^2 y)}{\partial z} = \cos^2 y + 0 + \sin^2 y = 1$$ $$\iiint \operatorname{div} \vec{A} \, dx \, dy \, dz = \iiint 1 \cdot dx \, dy \, dz = \begin{pmatrix} x = r \sin \theta \cos \varphi & z = r \cos \theta \\ y = r \sin \theta \sin \varphi & J = r^2 \sin \theta \end{pmatrix} = \int_{\varphi=0}^{2\pi} \int_{\theta=0}^{\pi} \int_{r=0}^{4} r^2 \sin \theta \, d\varphi \, d\theta \, dr = \int_{\varphi=0}^{2\pi} d\varphi \int_{\theta=0}^{\pi} \sin \theta \, d\theta \int_{r=0}^{4} r^2 \, dr = \varphi \big|_{0}^{2\pi} \cdot [-\cos \theta] \big|_{0}^{\pi} \cdot \frac{r^3}{3} \big|_{0}^{4} = 2\pi \cdot 2 \cdot \frac{4^3}{3} = \frac{256\pi}{3}$$

Answer:

$$\frac{256\pi}{3}$$