# Find the flux of \vec{F} across S where \vec{F}(x) y, z) = 6z\vec{i} + 2x\vec{j} + y\vec{k} and S...

## Question:

Find the flux of {eq}\vec{F} {/eq} across S where {eq}\vec{F}(x) y, z) = 6z\vec{i} + 2x\vec{j} + y\vec{k} {/eq} and S is the part of the plane 2x + 3y + 6z = 12 in the first octant; \vec{n} points upward.

## Flux through a surface:

Integration has various important applications. Double integrals are used to find the flux of a vector field through a given surface S. Find the normal to the given surface and equations of the surface to find the limits of integration. It is given by the formula as Flux= {eq}\int \int F\cdot \hat{n}dS {/eq}

S is the surface given by {eq}2x+3y+6z=12 {/eq}

Normal vector to the surface is given by {eq}n=\langle 2,3,6 \rangle {/eq}

Unit vector {eq}\hat{n}=\frac{1}{7}\langle 2,3,6 \rangle {/eq}

Now {eq}6z=12-2x-3y {/eq}

Calculating partial derivatives

{eq}z_{x}= -2, z_{y}= -3 {/eq}

{eq}dS=\sqrt{z_{x}^2+z_{y}^2+36}dA {/eq}

{eq}dS=\sqrt{4+9+36}=\sqrt{49}dA= 7 dA {/eq}

The region is bounded by the points {eq}(6,0,0), (0,4,0) {/eq} and {eq}(0,0,2) {/eq}

Thus flux is given by

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} \int_{0}^{4-\frac{2x}{3}} \langle 6z,2x,y \rangle \cdot \ \frac{1}{7}\langle 2,3,6 \rangle 7 dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} \int_{0}^{4-\frac{2x}{3}} \langle 6z,2x,y \rangle \cdot \ \langle 2,3,6 \rangle dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} \int_{0}^{4-\frac{2x}{3}} (12z+6x+6y)dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} \int_{0}^{4-\frac{2x}{3}} (24-4x-6y+6x+6y)dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} \int_{0}^{4-\frac{2x}{3}} (24-2x)dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} (24y-2xy)_{0}^{4-\frac{2x}{3}} dx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{6} (96-24x+\frac{4x^2}{3})dx {/eq}

{eq}\int \int F\cdot \hat{n}dS= (96x-12x^2+\frac{4x^3}{9})_{0}^{6} {/eq}

{eq}\int \int F\cdot \hat{n}dS= (576-432+96) {/eq}

{eq}\int \int F\cdot \hat{n}dS= 240 {/eq}