# Line integral of the second kind

1. May 11, 2015

### nuuskur

1. The problem statement, all variables and given/known data
Given the polar curve $r = 3\sqrt{\cos{2\varphi}},\ -\frac{\pi}{4}\leq\varphi\leq\frac{\pi}{4}$. Find the area of the surface enclosed by the curve using line integral of the second kind.

2. Relevant equations

3. The attempt at a solution
According to Green's theorem: if $F(x,y)$ and $G(x,y)$ are continous on a smooth closed graph ($r(\varphi)$ in this case), then $\oint_L F(x,y)\mathrm{d}x + G(x,y)\mathrm{d}y = \iint_D G_x(x,y) - F_y(x,y)\mathrm{d}x\mathrm{d}y$.

The textbook, however, is very ineloquent as to what exactly $F$ and $G$ are and there are no examples given with polar curves. Do I have to parameterize the curve? I don't understand.

Last edited: May 11, 2015
2. May 11, 2015

### Svein

This looks like a slightly offbeat version of Stokes' theorem.
This is the curve. Observe:
1. x = r*cos(φ), y = r*sin(φ)
2. sin(-φ) = -sin(φ), so the part above the x-axis is equal to the part below
3. You do not need any fancy theorems to do this integral.

3. May 11, 2015

### Ray Vickson

The area is $\iint_D 1 \, dx dy$, so you need to devise $F(x,y), G(x,y)$ that satisfy $G_x - F_y = 1$ everywhere (or, at least, inside the curve). Can you think of some convenient functions $F,G$ that have that property?

4. May 11, 2015

### Zondrina

The area of the region is what you mean.

The area of a polar region $r = f(\theta)$ is given by:

$$\frac{1}{2} \int_a^b f^2(\theta) \space d\theta$$

This will allow you to solve the question quite easily as you already have the limits.

On a side note, I think you are referring to the reverse implication of Green's theorem, which allows you to compute areas. Since the area of a region $D$ is given by:

$$\text{Area}(D) = \iint_D \space dA$$

We wish to choose the functions $P$ and $Q$ satisfying Green's theorem such that:

$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$$

There are several possibilities for choosing these functions, which all lead to the same result for the area of a region $D$. Lets choose $P = -\frac{1}{2}y$ and $Q = \frac{1}{2}x$. The following result can then be deduced directly from Green's theorem:

$$\text{Area}(D) = \frac{1}{2} \oint_C xdy - y dx = \iint_D \space dA$$

If you were to "go backwards" from $r = f(\theta)$ by using polar co-ordinates, you could also use the above result to get the exact same answer, i.e:

$$\text{Area}(D) = \frac{1}{2} \oint_C xdy - y dx = \frac{1}{2} \int_a^b f^2(\theta) \space d\theta = \iint_D \space dA$$

5. May 13, 2015

### nuuskur

Would the area then be
$$A = \int_{ -\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{r(\varphi)} r\mathrm{d}r\mathrm{d}\varphi$$
Edit: Yes, it is, thanks, I get it now.

Last edited: May 13, 2015