# Is this line integral computation correct? (Green's Thm application)

• Fractal20
In summary, the homework statement says that you are to find the value of the line integral \int_C(e-x^3-3y)dx+. The Attempt at a Solution says that this line integral is a double integral and that you need to find the limits of integration with respect to r.
Fractal20

## Homework Statement

Find the value of the line integral

$\int$ C(e-x^3 - 3y)dx + (tan y + y4 + x) dy

where C is the counterclockwise-oriented circle of radius 4 centered at (0,2).

## The Attempt at a Solution

First off, I didn't think this was path independent since the derivative of the dx term in respect to y equals -3 which doesn't equal the derivative of the dy term with respect to x, 1.

After getting stuck on some direct approaches, I realized it is probably a Green's Thm application.

This then translates to $\int$$\int$ 4 dA. So in polar coordinates the path of the circle is

x = 4 cos $\vartheta$
y = 4 sin $\vartheta$ + 2

This is where I get unsure. To try and get the limits on r I tried to say r = sqrt (x + y) and then plugged in the paramatized equations for x and y. This simplified to r = sqrt (20 + 16 sin $\vartheta$). So then I thought that r varies from 2 to this general thing and hence

$\int$ $^{2pi}_{0}$ $\int$$^{\sqrt{20 + 16 sin\vartheta}_{2}}$ 4 r dr d$\vartheta$.

Is this right? Note that the limits with respect to r is suppose to be 2 to that sqrt expression, the two just kept popping up a little below and to the right... In particular is this a correct way to find the respective limits of integration? Thanks!

## The Attempt at a Solution

Last edited:
Fractal20 said:
$\int$ C(e-x^3 - 3y)dx = (tan y + y4 + x) dy

Check that you've typed the integral correctly, as it doesn't make much sense as it is.

Did you mean:
$$\int_C (e^{-x^3}-3y)dx+(\tan y + y^4+x)dy$$

1. The radius of the circle is given directly in the problem statement.

2. The double integral of a constant is equal to the constant multiplied by the area of the domain of integration, which is a circle of a known radius here.

Oops, sorry that was suppose to be a + rather than =. Thanks Voko, I should have seen that. That clears it all up for me.

## 1. What is a line integral?

A line integral is a type of integral that is used to calculate the total value of a function along a given curve or path.

## 2. What is Green's Theorem and how is it applied?

Green's Theorem is a mathematical theorem that relates the line integral of a two-dimensional vector field to a double integral over the region enclosed by the curve. It is used to simplify the calculation of line integrals by converting them into double integrals.

## 3. How do you know if a line integral computation using Green's Theorem is correct?

To determine if a line integral computation using Green's Theorem is correct, you should first check that the vector field and the curve are both well-defined. Then, you should verify that the integral has been set up correctly and that all the necessary steps have been followed accurately. Lastly, you should compare the result to other known solutions to ensure it is correct.

## 4. What are the common mistakes made when using Green's Theorem to compute line integrals?

Some common mistakes when using Green's Theorem include setting up the integral incorrectly, using the wrong limits of integration, forgetting to include a negative sign, and making errors in the calculation of the double integral.

## 5. Can Green's Theorem be used to calculate line integrals in three dimensions?

No, Green's Theorem can only be applied in two dimensions. In three dimensions, the equivalent theorem is called Stokes' Theorem, which relates the surface integral of a three-dimensional vector field to a triple integral over the region enclosed by the surface.

• Calculus and Beyond Homework Help
Replies
10
Views
400
• Calculus and Beyond Homework Help
Replies
20
Views
437
• Calculus and Beyond Homework Help
Replies
12
Views
972
• Calculus and Beyond Homework Help
Replies
3
Views
220
• Calculus and Beyond Homework Help
Replies
8
Views
859
• Calculus and Beyond Homework Help
Replies
1
Views
475
• Calculus and Beyond Homework Help
Replies
4
Views
995
• Calculus and Beyond Homework Help
Replies
12
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
663
• Calculus and Beyond Homework Help
Replies
8
Views
746