# Use Stokes' Theorem to evaluate \int_CF \space dr , where C is oriented counterclockwise as...

## Question:

Use Stokes' Theorem to evaluate {eq}\int_CF \space dr {/eq}, where C is oriented counterclockwise as viewed from above.

{eq}F(x, y, z) = (x + y^2)i + (y + z^2)j + (z + x^2)k {/eq},

C is the triangle with vertices (5, 0, 0), (0, 5, 0), and (0, 0, 5).

## Stokes' theorem:

Stokes' theorem states that, given a closed curve C, the circulation of a vector field F equals the flow of its rotational through an arbitrary surface S with C as the edge, and oriented according to the rule of the right hand.

$$C=\int_C F= \iint_S (rot F) \, dS $$

## Answer and Explanation:

**Find the curl of the vector field F.**

{eq}\vec F = (x + y^2)i + (y + z^2)j + (z + x^2)k\\ \vec F = P \vec i +Q \vec j + R \vec k\\ curl F= \begin{pmatrix}\vec i&\vec j&\vec k\\ \frac {\partial}{\partial x}&\frac {\partial}{\partial y}&\frac {\partial}{\partial z}\\P&Q&R\end{pmatrix}\\ {/eq}

{eq}curl F= (\frac {\partial R}{\partial y} - \frac {\partial Q}{\partial z}, \frac {\partial P}{\partial z} - \frac {\partial R}{\partial x}, \frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}) \\ {/eq}

{eq}curl F= (0-2z,0-2x,0-2y)\\ curl F= (-2z,-2x,-2y)\\ {/eq}

**Find the equation of the plane that joins the given points**

{eq}A(5, 0, 0), B(0, 5, 0), and \,\,\, C(0, 0, 5) {/eq}

**To calculate the equation of the plane we use the formula:**

{eq}det \begin{pmatrix}x-x_a&y-y_a&z-z_a&\\x_b-x_a&y_b-y_a&z_b-z_a\\ x_c-x_a&y_c-y_a&z_c-z_a \end{pmatrix}=0\\ {/eq}

{eq}det \begin{pmatrix}x-5&y&z&\\-5&5&0\\-5&0&5\end{pmatrix}=0\\ {/eq}

{eq}x+y+z=5\\ {/eq}

**Applying the Stokes' Theorem.**

{eq}I=\int_C F= \iint_S (rot F) \, dS \\ \vec F = (x + y^2)i + (y + z^2)j + (z + x^2)k\\ curl F= (-2z,-2x,-2y) {/eq}

**Applying rectangular coordinates.**

{eq}r(y,z)=\left\{ \begin{array}{ll} x=x \\ y=y \,\,\,\,\,\, 0\leq x \leq 5\\ z=5-x-y \,\,\,\,\,\, 0 \leq y \leq 5-x\\ \end{array} \right. {/eq}

**Calculate the fundamental vector product.**

{eq}r_y \, x \, r_z = \begin{pmatrix}\vec i&\vec j&\vec k\\ 1 & 0 & -1 \\ 0 & 1& -1 \end{pmatrix}\\ r_y \, x \, r_z = 1 \vec i +0 \vec j +0 \vec k\\ r_y \, x \, r_z= (1, 1, 1) {/eq}

**Calculate the integral.**

{eq}(r_y \, x \, r_z) \cdot (rot F)=(1, 1, 1) \cdot (-2z,-2x,-2y)\\ (r_y \, x \, r_z) \cdot (rot F)= -2z,-2x,-2y\\ (r_y \, x \, r_z) \cdot (rot F)= -2(5-x-y),-2x,-2y\\ (r_y \, x \, r_z) \cdot (rot F)= 10\\ C=\int_C F= \int_{0}^{5} \int_{0}^{5-x} 10 \, dydx \\ C= \int_{0}^{5} \left. 10y \right|_{0}^{5-x} \, dx \\ C= \int_{0}^{5} 50-10x \, dx \\ C= \left. 50x -5x^2 \right|_{0}^{5}\\ C= 125 {/eq}

**Answer**

**The result of the integral is:** {eq}\displaystyle \Longrightarrow \boxed{125} {/eq}

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from UPSEE Paper 1: Study Guide & Test Prep

Chapter 8 / Lesson 10