# Line integral, incorrect setup

• jonroberts74
In summary, the given conversation involves determining the line integral of a vector field over a given path, which is parametrized and then integrated using the dot product. The solution to the problem is also discussed, with a potential mistake being pointed out and corrected.
jonroberts74

## Homework Statement

##\int_\mathscr{C} \vec{F}(\vec{r})\cdot d\vec{r}; \vec{F}(x,y,z) = <sin z, cos \sqrt{y}, x^3>## I am assuming ##\vec{r}## is the usual ##\vec{c}## used, so maybe this is where I am incorrect

## The Attempt at a Solution

C goes from (1,0,0) to (0,0,3)

Parametrizing C

##\mathscr{C}: \vec{c}(t) = (1-t)<1,0,0> + t<0,0,3> = <1-t, 0 ,3t>; 0 \le t \le 1 ##

##\vec{c}\,\,'(t) = <-t, 0, 3>##

##\vec{F}(\vec{c}(t) = <\sin 3t, 1, (1-t)^3>##

##\displaystyle \int_{0}^{1} <\sin 3t, 1, (1-t)^3> \cdot <-t, 0, 3>dt##

##\displaystyle \int_{0}^{1} -t \sin 3t + 0 + 3(1-t)^3 dt##

I got this far and integrated it but got the wrong answer, I checked my integration already so I integrated this setup correctly but I screwed up on the setup somewhere.

Check ##c'(t)## again.

1 person
Check ##c'(t)## again.

yeah I did the same mistake as the last one. thanks

Does the problem only say "C goes from (1,0,0) to (0,0,3)" or does it specifically say "the straight line from (1, 0, 0) to (0, 0, 3)?

HallsofIvy said:
Does the problem only say "C goes from (1,0,0) to (0,0,3)" or does it specifically say "the straight line from (1, 0, 0) to (0, 0, 3)?

says line segment from (1,0,0) to (0,0,3)

jonroberts74 said:

## Homework Statement

##\int_\mathscr{C} \vec{F}(\vec{r})\cdot d\vec{r}; \vec{F}(x,y,z) = <sin z, cos \sqrt{y}, x^3>## I am assuming ##\vec{r}## is the usual ##\vec{c}## used, so maybe this is where I am incorrect

## The Attempt at a Solution

C goes from (1,0,0) to (0,0,3)

Parametrizing C

##\mathscr{C}: \vec{c}(t) = (1-t)<1,0,0> + t<0,0,3> = <1-t, 0 ,3t>; 0 \le t \le 1 ##

##\vec{c}\,\,'(t) = <-t, 0, 3>##

##\vec{F}(\vec{c}(t) = <\sin 3t, 1, (1-t)^3>##

##\displaystyle \int_{0}^{1} <\sin 3t, 1, (1-t)^3> \cdot <-t, 0, 3>dt##

##\displaystyle \int_{0}^{1} -t \sin 3t + 0 + 3(1-t)^3 dt##

I got this far and integrated it but got the wrong answer, I checked my integration already so I integrated this setup correctly but I screwed up on the setup somewhere.
What do you get for the answer?

## What is a line integral?

A line integral is a mathematical concept used in calculus to measure the total value of a function along a specific path or curve.

## How is a line integral calculated?

A line integral is calculated by breaking down the path or curve into small segments and approximating the value of the function at each point. These values are then added together to find the total value of the line integral.

## What is the purpose of a line integral?

The purpose of a line integral is to calculate the total value of a function along a specific path or curve. This can be useful in many scientific and engineering applications, such as calculating work done by a force or finding the mass of a curved object.

## What is an incorrect setup for a line integral?

An incorrect setup for a line integral is when the path or curve is not properly broken down into small segments, or when the values of the function are not properly approximated at each point. This can lead to inaccurate results.

## How can an incorrect setup affect the accuracy of a line integral?

An incorrect setup for a line integral can greatly affect the accuracy of the results. If the path or curve is not properly divided into small segments, or if the values of the function are not properly approximated, the total value of the line integral will be incorrect. This can lead to incorrect conclusions and potentially impact the validity of any scientific or engineering calculations based on the line integral.

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