Question

6.) Use the Divergence theorom to evaluate the surface integral Sts Fods for the given vector field F and the oriented surface S. For closed surfaces, use the positive lost ward) orientation, F(x, y, z) = -xi - y jt zh, s is the part of 2 = x²ty² under the planes 2 = 4 with up ward orientation.

Answer 1

The divergence theorem states that

F. ds = SIJ, (5. F) av dV (1) JJS

is the volume enclosed by the closed surface . The given vector field is

$\vec{F} = -x\hat{i}-y\hat{j}+z\hat{k}\Rightarrow \nabla\cdot\vec{F} = -\frac{\partial x}{\partial x}-\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z} = -1 \quad (2)$

The volume integral becomes

(8) ЛР - Л. АР (4-д) JГ

To calculate the volume integral we will work in cylindrical coordinates

$\rho = \sqrt{x^{2}+y^{2}},\; \phi = \arctan\left(\frac{y}{x} \right ),\; z = z \quad (4)$

The volume element of cylindrical coordinates is

$dV = \rho d\rho d\phi dz \quad (5)$

The equation for the surface in cylindrical coordinates is

$z = \rho^{2}\Rightarrow \rho = \sqrt{z} \quad (6)$

Hence

$\iiint_{V}\left(\nabla\cdot\vec{F} \right )dV = -\int_{z=0}^{4}dz\int_{0}^{\sqrt{z}}\rho d\rho\int_{0}^{2\pi}d\phi = -2\pi\int_{0}^{4}\frac{z}{2}dz = -8\pi \quad (7)$

Using divergence theorem, we get

$\iint_{S}\vec{F}\cdot\vec{dS} = -8\pi \quad (8)$