# Let \vec F(x,y,z) = 2\vec i + 3\vec j + \vec k and let S be a 3 \times 3 square on the plane 4x +...

## Question:

Let {eq}\vec F(x,y,z) = 2\vec i + 3\vec j + \vec k {/eq} and let {eq}S {/eq} be a {eq}3 \times 3 {/eq} square on the plane {eq}4x + y + 2z = 8 {/eq} oriented in the direction of increasing {eq}z {/eq}. Find the flux of {eq}\vec F {/eq} over {eq}S {/eq}.

## Flux through a surface:

Dot product and double integrals both collectively are used to find the flux of a vector field through a given surface S. First, 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}4x+y+2z=8 {/eq}

Normal vector to the surface is given by {eq}n=\langle 4,1,2 \rangle {/eq}

Unit vector {eq}\hat{n}=\frac{1}{\sqrt{21}}\langle 4,1,2 \rangle {/eq}

Now {eq}2z=8-4x-y {/eq}

Calculating partial derivatives

{eq}z_{x}= -4, z_{y}= -1 {/eq}

{eq}dS=\sqrt{z_{x}^2+z_{y}^2+4}dA {/eq}

{eq}dS=\sqrt{16+1+4}=\sqrt{21}dA {/eq}

The function is given by

{eq}F=\langle 2,3,1 \rangle {/eq}

The region is bounded by the points {eq}(2,0,0), (0,8,0) {/eq} and {eq}(0,0,4) {/eq}

Thus flux is given by

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{2} \int_{0}^{8-4x} \langle 2,3,1 \rangle \cdot \ \frac{1}{\sqrt{21}}\langle 4,1,2 \rangle \sqrt{21}dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{2} \int_{0}^{8-4x} \langle 2,3,1 \rangle \cdot \langle 4,1,2 \rangle dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{2} \int_{0}^{8-4x} 13dydx {/eq}

{eq}\int \int F\cdot \hat{n}dS=\int_{0}^{2} 13(8-4x) dx {/eq}

{eq}\int \int F\cdot \hat{n}dS=13 \int_{0}^{2}(8-4x)dx {/eq}

{eq}\int \int F\cdot \hat{n}dS=13 (8x-2x^2)_{0}^{2} {/eq}

{eq}\int \int F\cdot \hat{n}dS=13 (16-8) {/eq}

{eq}\int \int F\cdot \hat{n}dS=104 {/eq}