# Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use. F = <<x sin y, -cos y, z sin y>> , S is the boundary of the region bounded by the planes x = 1, y = 0, y = pi/2, z = 0, and z = x.

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$F=⟨x\mathrm{sin}y,-\mathrm{cos}y,z\mathrm{sin}y⟩$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $y=\frac{\pi }{2},z=0$, and z = x.
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odgovoreh
Step 1
Given:
$F=⟨x\mathrm{sin}y,-\mathrm{cos}y,zz\mathrm{sin}y⟩$, the surface S is the boundary of the region bounded by the planes x = 1, $y=\frac{\pi }{2},z=0$, and z = x.
Step 2
Using Divergence theorem
$\int {\int }_{D}\int \mathrm{\nabla }\cdot Fdv=\int {\int }_{S}F\cdot nds$
Compute The divergence of the field
$\mathrm{\nabla }\cdot F=\frac{\partial }{\partial x}\left(x\mathrm{sin}y\right)+\frac{\partial }{\partial y}\left(-\mathrm{cos}y\right)+\frac{\partial }{\partial z}\left(z\mathrm{sin}y\right)$
$=\mathrm{sin}y\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(-\mathrm{cos}y\right)+\mathrm{sin}y\frac{\partial }{\partial z}\left(z\right)$
$=\mathrm{sin}y+\mathrm{sin}y+\mathrm{sin}y$
$=3\mathrm{sin}y$
Step 3
Therefore, the outward flux is
$\int {\int }_{D}\int \mathrm{\nabla }\cdot Fdv={\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{1}{\int }_{0}^{x}3\mathrm{sin}ydzdxdy$
$=3{\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{1}\mathrm{sin}y{\left[z\right]}_{0}^{x}dxdy$
$=3{\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{1}x\mathrm{sin}ydxdy$
$=3{\int }_{0}^{\frac{\pi }{2}}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}\mathrm{sin}ydy$
$=\frac{3}{2}{\int }_{0}^{\frac{\pi }{2}}\left[\mathrm{sin}ydy$
$=\frac{3}{2}{\int }_{0}^{\frac{\pi }{2}}\left[\mathrm{sin}ydy$
$=\frac{3}{2}{\left[-\mathrm{cos}y\right]}_{0}^{\frac{\pi }{2}}$
$=\frac{3}{2}\left(-\mathrm{cos}\left(\frac{\pi }{2}\right)-\left(-\mathrm{cos}\left(0\right)\right)\right)$