# Use Stokes' Theorem to find the circulation of \vec F = 2y\hat i + 5z\hat j + 5x\hat k around the...

## Question:

Use Stokes' Theorem to find the circulation of {eq}\vec F = 2y\hat i + 5z\hat j + 5x\hat k {/eq} around the triangle obtained by tracing out the path {eq}(6, 0, 0) {/eq} to {eq}(6, 0, 6) {/eq} to {eq}(6, 5, 6) {/eq} back to {eq}(6, 0,0). {/eq}

## Application of Stoke's Theorem:

Rotational is understood as the vector operator that shows the tendency of a field to induce rotation around a point. Mathematically it is expressed as the circulation of a vector field on a closed path.

## Answer and Explanation:

Given that, {eq}\vec F = 2y\hat i + 5z\hat j + 5x\hat k {/eq} around the triangle obtained by tracing out the path {eq}(6, 0, 0) {/eq} to {eq}(6, 0, 6) {/eq} to {eq}(6, 5, 6) {/eq} back to {eq}(6, 0,0). {/eq}

As per Stokes theorem,

{eq}\int _L \overrightarrow{F} \cdot dr = \int \int_S curl\overrightarrow{F}\cdot \widehat{n}ds {/eq}

then,

{eq}curl\overrightarrow{F} = \begin{vmatrix} \overrightarrow{i} &\overrightarrow{j} &\overrightarrow{k} \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ 2y & 5z & 5x \end{vmatrix}\\ curl\overrightarrow{F} = \overrightarrow{i}(\frac{\partial }{\partial y}(5x)-\frac{\partial }{\partial z}(5z))-\overrightarrow{j}(\frac{\partial }{\partial x}(5x)-\frac{\partial }{\partial z}( 2y))+\overrightarrow{k}(\frac{\partial }{\partial x}(-5z)-\frac{\partial }{\partial y}(2y))\\ curl\overrightarrow{F} = \overrightarrow{i}(-5)-\overrightarrow{j}(5)+\overrightarrow{k}(-2)\\ curl\overrightarrow{F} = -5\overrightarrow{i}-5\overrightarrow{j}-2\overrightarrow{k}\\ \int _L \overrightarrow{F} \cdot dr = \int \int_S (-5\overrightarrow{i}-5\overrightarrow{j}-2\overrightarrow{k})\cdot (\overrightarrow{i}+0\overrightarrow{j}+0\overrightarrow{k})dS\\ \int _L \overrightarrow{F} \cdot dr = \int \int_S (-5)dS\\ {/eq}

where S is the surface bounded by (0,0),(0,6), (5,6) and S represent the triangular region with {eq}y=b=5 \ and \ z=h=6 {/eq}

{eq}\int _L \overrightarrow{F} \cdot dr = (-5) * \ Area \ of \ the \ triangle \ in \ S\\ \int _L \overrightarrow{F} \cdot dr = (-5) *\frac{1}{2}(5)(6)\\ \int _L \overrightarrow{F} \cdot dr = -75 {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14