# Vector Calculus Homework: Evaluate \int F dr

• mamma_mia66
In summary: I then found the derivative of the potential function with respect to x and y. I then used the fundamental theorem of line integral to find the derivative of the derivative with respect to x and y. I then used the chain rule to find the derivative of the derivative with respect to x and y. Lastly, I used the above information to find the final value of the function.
mamma_mia66

## Homework Statement

Evaluate $$\int$$ F dr
if F(x,y) =(6x2 +4y) i + (4x-2y) j and the curve C is a smooth curve from (1,1) to (2,3).

## The Attempt at a Solution

I took partial derivatives with respect of y for the first term and with respect of x for the second term.
4=4 => F is conservative, but from now I am confused how to solvet. I need at least the idea.

I can find the patential function and the plug in the points values (1,1) and (2,3) for y and x .

I am not sure if I can apply the Fundamental Thm of Line Integrals.

I will appreciate any ideas how to finish the problem. Thank you.

Hi mamma_mia66!
mamma_mia66 said:
Evaluate $$\int$$ F dr
if F(x,y) =(6x2 +4y) i + (4x-2y) j and the curve C is a smooth curve from (1,1) to (2,3).

4=4 => F is conservative, but from now I am confused how to solvet. I need at least the idea.

Since you've proved F is conservative, that means you're free to choose any path …

so just choose whatever looks easiest …

I'd go for either (1,1) to (1,3) to (2,3), or (1,1) to (2,1) to (2,3)

Saying that Fdr is conservative means that there exist a function f(x,y) such that df= F ds and so the integral of Fdr between two points $(x_1,y_1)$ and $(x_2,y_2)$ is just $f(x_2,y_2)- f(x_1,y_1)$

So another way to do this is to find f(x,y) such that
$$df= \frac{\partial f}{dx}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}= F dr$$
evaluated at (2,3) minus the value at (1, 1).
That means you must have
$$\frac{\partial f}{\partial x}= 6x^2+ 4y$$
and
$$\frac{\partial f}{\partial y}= 4x- 2y[/itex] From the first, f(x,y) must equal $2x^2+ 4xy+ g(y)$ since the "constant of integration" must depend on y only. Now, knowing that [tex]\frac{\partial f}{\partial y}= \frac{\partial (2x^2+ 4xy+ g(y)}{\partial y}= 4x+ \frac{dg}{dy}= 4x- 2y$$
what must f equal?

Last edited by a moderator:
Thank you so much . I did finish the problem. I found the potential function and I pluged in the given points for y and x.

## 1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with the study of vectors, which are quantities that have both magnitude and direction. It involves the use of vectors to solve problems in calculus, such as finding derivatives and integrals in multiple dimensions.

## 2. What is the purpose of evaluating \int F dr?

The purpose of evaluating \int F dr is to find the total work done by a vector field F on a particle as it moves along a given path represented by the curve r. This involves finding the line integral of the vector field along the given curve.

## 3. How do you evaluate \int F dr?

To evaluate \int F dr, you first need to parametrize the given curve r(t) and then find the dot product between the vector field F and the tangent vector r'(t). This dot product is then integrated with respect to t, from the initial value of t to the final value of t.

## 4. What are some common applications of evaluating \int F dr?

Evaluating \int F dr has many applications in physics and engineering, such as calculating the work done by a force on an object, finding the circulation of a fluid flow, and determining the flux of a vector field through a surface. It is also used in fields like computer graphics and robotics.

## 5. Are there any tips for simplifying the process of evaluating \int F dr?

One tip for simplifying the process of evaluating \int F dr is to choose a parametrization that makes the integrand simpler to work with. It is also helpful to break the given curve into smaller segments and evaluate the integral for each segment separately. Additionally, using symmetry or known properties of the vector field can also simplify the evaluation process.

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