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Calculus 3 : Line Integrals of Vector Fields

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Example Questions

Example Question #3832 : Calculus 3

Evaluate $\int_C \vec{F}\cdot d\vec{r}$ , where $\vec{F}(x,y,z)=10x^2y^2z^3\vec{i}+5z\vec{j}+2yz\vec{k}$ , and is the curve given by $\vec{r}(t)=t\vec{i}+t^2\vec{j}+t^3\vec{k}$ , $0\leq t \leq 1$ .

Possible Answers:

$\frac{24}{67}$

$\frac{59}{24}$

$\frac{13}{24}$

$\frac{67}{24}$

Correct answer:

$\frac{67}{24}$

Explanation:

First we need to evaluate the vector field evaluated along the curve.

$\vec{F}(\vec{r}(t))=10t^2(t^2)^2(t^3)^3\vec{i}+5t^3\vec{j}+2t^2t^3\vec{k}$

$\vec{F}(\vec{r}(t))=10t^{15}\vec{i}+5t^3\vec{j}+2t^5\vec{k}$

Now we need to find the derivative of $\vec{r}(t)$

$\vec{r}\ '(t)=\vec{i}+2t\vec{j}+2t^2\vec{k}$

Now we can do the product of $\vec{F}(\vec{r}(t))$ and $\vec{r}(t)$ .

$\vec{F}(\vec{r}(t))\cdot \vec{r}(t)=(10t^{15}, 5t^3, 2t^5)\cdot(1, 2t, 2t^2)$

$=10t^{15}+10t^5+4t^7$

Now we can put this into the integral and evaluate it.

$\int_{0}^{1} 10t^{15}+10t^5+4t^7 \ dt$

$=\Big(\frac{5}{8}t^{16}+\frac{5}{3}t^6+\frac{1}{2}t^8\Big)\Big|_{0}^{1}$

$=\Big(\frac{5}{8}(1)^{16}+\frac{5}{3}(1)^6+\frac{1}{2}(1)^8\Big)$

$=\frac{5}{8}+\frac{5}{3}+\frac{1}{2}=\frac{13}{6}+\frac{5}{8}=\frac{30}{48}+\frac{104}{48}=\frac{134}{48}=\frac{67}{24}$

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Example Question #2 : Line Integrals Of Vector Fields

Find the work done by a particle moving in a force field F = <x^2,y^2> , moving from (2,4) to (5,25) on the path given by y=x^2 .

Possible Answers:

5226

2613

Correct answer:

5226

Explanation:

The formula for work is given by

$W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$ .

Writing our path in parametric equation form, we have

x=t, y=t^2, 2<t<5 .

Hence

$\mathbf{r}(t) = <t,t^2>, F(\mathbf{r}(t))=<t^2,t^4>, \mathbf{r}'(t)= <1,2t>$

Plugging this into our work equation, we get

$W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$

$= \int_{2}^{5}<t^2,t^4> \cdot <1,2t> dt$

$= \int_2^5 t^2+2t^5 dt$ .

= 5226

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Example Question #3831 : Calculus 3

Evaluate $\int \vec{F}\cdot d\vec{r}$ on the curve $\vec{r}(t)=\left \langle 2t, 3t, t^3\right \rangle$ , $0\leq t \leq 1$ , where $\vec{F}=\left \langle 2xz, y^2, z^2\right \rangle$ .

Possible Answers:

$\frac{164}{15}$

$\frac{32}{6}$

$\frac{32}{3}$

Correct answer:

$\frac{164}{15}$

Explanation:

The line integral of a vector field is given by

$\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt$

So, we must evaluate the vector field on the curve:

$\vec{F}(\vec{r}(t))=\left \langle 2(2t)(t^3), 9t^2, t^6\right \rangle$

Then, we take the derivative of the curve with respect to t:

$\vec{r'}(t)=\left \langle 2, 3, 3t^2\right \rangle$

Taking the dot product of these two vectors, we get

8t^4+27t^2+3t^8

This is the integrand of our integral. Integrating, we get

$\int_{0}^{1}(8t^4+27t^2+3t^8 )dt = \frac{8t^5}{5}+\frac{27t^3}{3}+\frac{3t^9}{9}\Big|^{1}_{0}= \frac{164}{15}$

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Example Question #4 : Line Integrals Of Vector Fields

Evaluate the integral $\int \vec{F}\cdot d\vec{r}$ on the curve $\vec{r}(t)=\left \langle t^3, t^2, t+2\right \rangle$ , where $\vec{F}=\left \langle xz, yz, xy\right \rangle$ , on the interval $0 \leq t \leq 1$

Possible Answers:

$\frac{629}{105}$

$\frac{419}{210}$

$\frac{629}{210}$

Correct answer:

$\frac{629}{210}$

Explanation:

The line integral of the vector field is equal to

$\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt$

The parameterization (using the corresponding elements of the curve) of the vector field is

$\vec{F}(\vec{r}(t))= \left \langle t^4+2t^3, t^3+2t^2, t^5\right \rangle$

The derivative of the parametric curve is

$\vec{r'}(t)=\left \langle 3t^2, 2t, 1 \right \rangle$

Taking the dot product of the two vectors, we get

$\vec{F}(\vec{r}(t))\cdot \vec{r'(t)}= 3t^6+6t^5+2t^4+4t^3+t^5= 3t^6+7t^5+2t^4+4t^3$

Integrating this with respect to t on the given interval, we get

$\int_{0}^{1}(3t^6+7t^5+2t^4+4t^3)dt=\frac{3t^3}{7}+\frac{7t^6}{6}+\frac{2t^5}{5}+t^4\left.\begin{matrix} 1\\0 \end{matrix}\right|= \frac{629}{210}$

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Example Question #5 : Line Integrals Of Vector Fields

Calculate $\int \vec{F}\cdot d\vec{r}$ on the interval $0\leq t \leq1$ , where $\vec{F}=\left \langle x, xy, z^2\right \rangle$ and $\vec{r}(t)=\left \langle 2t^2, t+2, t\right \rangle$

Possible Answers:

$\frac{25}{6}$

$\frac{5}{3}$

$\frac{25}{3}$

Correct answer:

$\frac{25}{6}$

Explanation:

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.

The vector field evaluated on the given curve is

$\vec{F}(\vec{r}(t))=\left \langle 2t^2, 2t^3+4t^2, t^2\right \rangle$

The derivative of the curve is given by

$\vec{r'}(t)=\left \langle 4t, 1, 1\right \rangle$

The dot product of these is

$\vec{F}(\vec{r(t)})\cdot \vec{r'}(t)=11t^3+5t^2$

Integrating this over our given t interval, we get

$\int_{0}^{1}(10t^3+5t^2)dt=\frac{5t^4}{2}+\frac{5t^3}{3}\Big|^{1}_{0}= \frac{25}{6}$

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Calculus 3 : Line Integrals of Vector Fields

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3832 : Calculus 3

Example Question #2 : Line Integrals Of Vector Fields

Example Question #3831 : Calculus 3

Example Question #4 : Line Integrals Of Vector Fields

Example Question #5 : Line Integrals Of Vector Fields

All Calculus 3 Resources

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