# Polar vs. Rectangular Integral

I am working on a problem set that requires one to convert a polar equation into rectangular form, integrate using the rectangular equation and then integrate using the original polar equation.

The polar equation is:

$r = \frac{1}{1+\sin\theta}$, which I converted to:

$y = \frac{-1}{2}(x^2 - 1)$

I then computed the integral between -1 and 1:

$$\int_{-1}^{1}{\frac{-1}{2}(x^2-1)}\,dx=\frac{-1}{2}(\frac{x^3}{3} - x)$$

evaluated at 1 and -1, gives a value of $\frac{2}{3}$. No complaints.

However, when I integrate using the polar equation (using 0 and $\pi$ as my lower and upper limits, respectively), I keep ending up with $2$ as an answer.

$$\int_{0}^{\pi}{\frac{1}{1+\sin\theta}}$$

for which the antiderivative (one of them, at least) is

$$\tan\theta - \sec\theta$$ evaluated at 0 and $\pi$.

I have checked both of these with an online integral calculator, and keep coming up with the same results. Can anyone clarify the reason for this discrepancy? Or maybe my math is off . . .

Thanks !

That's because you're using the wrong expression to calculate the area in polar coordinates, it should be

$$\int_0^\pi \frac{1}{2}r^2{\rm d}\theta = \frac{1}{2}\int_0^{\pi}\frac{1}{(1 + \sin\theta)^2}{\rm d}\theta = -\left.\frac{1}{6}\frac{\cos(3\theta/2) - 3\sin(\theta/2)}{6[\cos(\theta/2) + \sin(\theta/2)]^3}\right|_{0}^{\pi} = \color{blue}{\frac{2}{3}}$$

• Many thanks for the re-steer in the right direction. How did you arrive at the antiderivative? I found a different one, using 1 +sin(theta) = 2cos^2(theta/2). – aupetitmoulin May 6 '18 at 15:55

The problem is that the area element in polar coordinates should be $\frac{1}{2}r^2 d\theta$: as the angle changes by $\delta\theta$, the area changes by roughly the area of the sector with radius $r$ and angle $\delta\theta$, which is $\frac{1}{2}r^2\delta\theta$ (this isn't quite right, since in fact $r(\theta+\delta\theta) = r(\theta)+\delta\theta r'(\theta)+\dotsb$, but the other contributions are higher-order and so disappear in the limit as $\delta\theta \to 0$ in the sum).