Line integral $\int\limits_L \left(3y+e^{cos(x)}\right)\ dx + \left(7x-\sqrt[3]{y^4+10}\right)\ dy$ for $x^2+y^2=3^2$

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I have attempted to calculate the line integral

$$\int\limits_L \left(3y+e^{\cos(x)}\right)\ dx + \left(7x-\sqrt[3]{y^4+10}\right)\mathrm{d}y$$

where $L$ is the circle $x^2+y^2=3^2$.

I decided to use Green's theorem by calculating the partial derivatives and subtracting.

$$\frac{\partial}{\partial x}\left(7x-\sqrt[3]{y^4+10}\right) = 7$$

$$\frac{\partial}{\partial y}\left(3y+e^{cos(x)}\right) = 3$$

$$\int\limits_L \left(3y+e^{cos(x)}\right)\ dx + \left(7x-\sqrt[3]{y^4+10}\right)\ dy = \iint\limits_D 4\ dx\ dy$$

It seemed to me at this point, that I should use polar coordinates since we are dealing with a circle.

I let $\theta$ be bounded by $0$ and $2\pi$ and $r$ be from $0$ to $3$ (the radius of the circle).

When doing change of variables, I let the Jacobian be $r$ to get:

$$\int_0^3 \int_0^{2\pi} 4r\ d\theta\ dr$$

Which comes out to $36\pi$ in the end.

Have I done the change of variables correctly?

Is there a way to check that I got the right answer?