Evaluate \int_C \overrightarrow{F} \cdot d\overrightarrow{r} when \overrightarrow{F}(x,y,z)=...


Evaluate {eq}\int_C \overrightarrow{F} \cdot d\overrightarrow{r}{/eq} when {eq}\overrightarrow{F}(x,y,z)= z\overrightarrow{i} + y\overrightarrow{j}+x^2\overrightarrow{k}{/eq} and {eq}C{/eq} is the path

{eq}r(t)= 3t\overrightarrow{i}+2t\overrightarrow{j}+t\overrightarrow{k}{/eq} for {eq}0 \leq t \leq 2{/eq}

Line integral across a curve :

Vector operations like dot product with little use of derivatives will be applied to solve line integral of a vector function across a curve C given in the parametric form. First put the values from parametric function {eq}r(t) {/eq} into {eq}F(x,y,z) {/eq} and then taking its dot product with {eq}r'(t) {/eq}. Finally integrate it with respect to t with given range of t as its limit

Answer and Explanation:

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

The parametric curve is given by

{eq}\vec r (t) = 3t \vec i + 2t \vec j+t \vec k {/eq}

Hence we get the values

{eq}x=3t, y=2t, z=t {/eq}

Writing the function as parametric curve

{eq}F(r(t))=t \vec i +2t \vec j+9t^2 \vec k {/eq}

{eq}\vec r'(t) = 3 \vec i +2 \vec j+ \vec k {/eq}

The integral is solved as

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r =\int_{a}^{b} F(r(t))\cdot \vec r'(t) dt {/eq}

As {eq}0 \leq t \leq 2; a=0, b=2 {/eq}

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r =\int_{0}^{2} (t \vec i +2t \vec j+9t^2 \vec k) \cdot ( 3 \vec i +2 \vec j+ \vec k) dt {/eq}

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r =\int_{0}^{2} [ 3t+4t+9t^2 ] dt=\int_{0}^{2} (9t^2+7t)dt {/eq}

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r=\left | 3t^3+\frac{7t^2}{2} \right |_{0}^{2} {/eq}

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r=3(8)+14 {/eq}

{eq}\displaystyle \int_{C} \vec F \cdot d \vec r=38 {/eq}

Learn more about this topic:

Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13

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