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

## 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:

DryRun
Gold Member
$\int$ 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.