# Compute the flux of the vector field \vec F(x, y, z) = 2x \vec j + 4y \vec k through the surface...

## Question:

Compute the flux of the vector field {eq}\vec F(x, y, z) = 2x \vec j + 4y \vec k {/eq} through the surface S, where S is the part of the surface {eq}z = -y + 2 {/eq}, above the rectangle {eq}0 \leq x \leq 5, 0 \leq y \leq 2 {/eq}, oriented upward.

flux = ?

## Integration

A mathematical quantity that represents the combine the differential form of data or variable function is known as integration. It helps in the physics discipline to find the point of the function or equation of the system.

Given Data:

• The vector field is: {eq}\mathop F\limits^ \to = 2x\mathop j\limits^ \to + 4y\mathop k\limits^ \to {/eq}
• {eq}z = - y + 2 {/eq}
• {eq}0 \le x \le 5 {/eq}
• {eq}0 \le y \le 2 {/eq}

The expression for surface by vector form is

{eq}\mathop r\limits^ \to \left( {x,y} \right) = x\mathop i\limits^ \to + y\mathop j\limits^ \to + z\mathop k\limits^ \to {/eq}

Substitute and solve the above expression

{eq}\mathop r\limits^ \to \left( {x,y} \right) = x\mathop i\limits^ \to + y\mathop j\limits^ \to + \left( { - y + 2} \right)\mathop k\limits^ \to \cdots\cdots\rm{(I)} {/eq}

Differentiate the above expression with respect to x

{eq}\begin{align*} \dfrac{{d\left( {\mathop r\limits^ \to \left( {x,y} \right)} \right)}}{{dx}} &= \dfrac{{d\left( {x\mathop i\limits^ \to + y\mathop j\limits^ \to + \left( { - y + 2} \right)\mathop k\limits^ \to } \right)}}{{dx}}\\ \mathop {{r_{ox}}}\limits^ \to &= \mathop i\limits^ \to \end{align*} {/eq}

Differentiate the expression (I) with respect to y

{eq}\begin{align*} \dfrac{{d\left( {\mathop r\limits^ \to \left( {x,y} \right)} \right)}}{{dy}} &= \dfrac{{d\left( {x\mathop i\limits^ \to + y\mathop j\limits^ \to + \left( { - y + 2} \right)\mathop k\limits^ \to } \right)}}{{dy}}\\ \mathop {{r_{oy}}}\limits^ \to &= \mathop j\limits^ \to - \mathop k\limits^ \to \end{align*} {/eq}

The expression for normal vector is

{eq}\mathop n\limits^ \to = \mathop {{r_{ox}}}\limits^ \to \times \mathop {{r_{oy}}}\limits^ \to {/eq}

Substitute and solve the above expression

{eq}\begin{align*} \mathop n\limits^ \to &= \mathop i\limits^ \to \times \left( {\mathop j\limits^ \to - \mathop k\limits^ \to } \right)\\ \mathop n\limits^ \to &= \begin{vmatrix} \mathop i\limits^ \to & \mathop j\limits^ \to & \mathop k\limits^ \to & \\ 1& 0 & 0\\ 0& -1 &1& \end{vmatrix} \end{align*} {/eq}

{eq}\begin{align*} \mathop n\limits^ \to &= \left( {0 \times \left( { - 1} \right) - 0\left( 0 \right)} \right)\mathop i\limits^ \to - \left( {1 \times \left( { - 1} \right) - 0\left( 0 \right)} \right)\mathop j\limits^ \to + \left( {1 \times \left( 1 \right) - 0\left( 0 \right)} \right)\mathop k\limits^ \to \\ &= 0 + \mathop j\limits^ \to + \mathop k\limits^ \to \\ &= \mathop j\limits^ \to + \mathop k\limits^ \to \end{align*} {/eq}

The area of small surface area is

{eq}dS = dxdy {/eq}

The expression for flux is

{eq}{\rm{Flux}} = \int {\int {\mathop F\limits^ \to } } .\mathop n\limits^ \to dS {/eq}

Substitute and solve the above expression with respective limit

{eq}\begin{align*} {\rm{Flux}} &= \int_0^2 {\int_0^5 {\left( {2x\mathop j\limits^ \to + 4y\mathop k\limits^ \to } \right)} } .\left( {\mathop j\limits^ \to + \mathop k\limits^ \to } \right)dxdy\\ &= \int_0^2 {\int_0^5 {\left( {2x + 4y} \right)} } dxdy\\ &= \int_0^2 {\left[ {\left( {\dfrac{{2{x^2}}}{2} + 4xy} \right)} \right]} _0^5dy\\ &= \int_0^2 {\left[ {\left( {{{\left( 5 \right)}^2} + 4\left( 5 \right)y - 0} \right)} \right]} _0^5dy\\ &= \int_0^2 {\left[ {25 + 20y} \right]} dy \end{align*} {/eq}

Integrate the above expression with respect to y

{eq}\begin{align*} {\rm{Flux}} &= \int_0^2 {\left[ {25 + 20y} \right]} dy\\ &= \left[ {25y + \dfrac{{20{y^2}}}{2}} \right]_0^2\\ &= \left[ {25\left( 2 \right) + 10{{\left( 2 \right)}^2} - 0} \right]\\ &= 90 \end{align*} {/eq}

Thus the flux of vector field is {eq}90 {/eq}