# Given a vector field F(x,y) = (6x^2 - 2y) \vec i + (8y^3 - 2x)\vec j . Use the fundamental...

## Question:

Given a vector field {eq}F(x,y) = (6x^2 - 2y) \vec i + (8y^3 - 2x)\vec j {/eq} .

Use the fundamental theorem of line integral to evaluate {eq}\int_C F\cdot dr {/eq} where C is a curve given by

{eq}r(t) = \left \langle t + 1, 3t \right \rangle ; 0 \leq t \leq 1 {/eq} .

## Line integrals:

Line integrals are useful in many branches, specifically in physics to calculate the work done by a force on a moving object. The line integral can be calculated as follows:

{eq}\int_C \vec F \cdot \vec dr = \int_ {a}^{b} \vec F (r(t)) \cdot r'(t) \, dt {/eq}

## Answer and Explanation:

**Calculate the curvilinear integral.**

{eq}r(t)=\left\{ \begin{array}{ll} x= t+1 \\ y= 3t \,\,\,\,\,\, 0\leq t \leq 1 \end{array} \right. {/eq}

**With this parameterization we have:**

{eq}F(x,y) = (6x^2 - 2y) \vec i + (8y^3 - 2x)\vec j\\ F(r(t))= (6(t+1)^2 -2(3t)) \vec i + (8(3t)^3 -2(t+1))\vec j\\ F(r(t))= (6(t^2+2t+1)-6t) \vec i + (216t^3 -2t-2)\vec j\\ F(r(t))= (6t^2+6t+6) \vec i + (216t^3 -2t-2)\vec j\\ r'(t)= \vec i +3 \vec j\\ F(r(t)) \cdot r'(t)= (6t^2+6t+6)+3(216t^3 -2t-2)\\ F(r(t)) \cdot r'(t)= 6t^2+6t+6+648t^3-6t-6\\ F(r(t)) \cdot r'(t)= 6t^2+648t^3 {/eq}

**Calculate the integral.**

{eq}C=\int_C \vec F \cdot \vec dr = \int_{0}^{1} F(r(t)) \cdot r'(t) \, dt\\ C= \int_{0}^{1} 6t^2+648t^3 \, dt\\ C= \left. 2t^3+162t^4 \right|_{0}^{1} \\ C=164 {/eq}

**Answer**

**The result of the line integral is:** {eq}\displaystyle \Longrightarrow \boxed{164} {/eq}

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from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2