Integral of a polar equation $r=1+\cos\theta$ from limits $-\pi/2$ to $0$ does not give me the same value as integrating from $3\pi/2$ to $0$. Why?

Question

To find the area of the cardioid (given by the polar equation $r=1+\cos(\theta)$)

I am trying to integrate a polar function ${r=1+\cos(\theta)}$, I do not have a problem with indefinite integration but if I were to integrate it from the limits $\frac{-\pi}{2}$ to $0$, I get the answer as: $\frac{\pi+2}{2}$

$$\int_{-\pi/2}^0 1+\cos(\theta)\ d\theta$$ $$=[\theta]^{0}_{-\pi/2}+[\sin(\theta)]^{0}_{-\pi/2}$$

$$=(0-(-\frac{\pi}{2}))+(\sin(0)-\sin(\frac{\pi}{2}))$$

$$=\frac{\pi+2}{2}$$ However, I have been assuming that if I were to integrate the same function from $\frac{3\pi}{2}$ to $0$, I would be getting the same as the previously obtained value (i.e. $\frac{\pi+2}{2}$). But this is not the case, I get a value of $-\frac{3\pi-2}{2}$:

$$\int_{3\pi/2}^0 1+\cos(\theta)\ d\theta$$

$$=[\theta]^{0}_{3\pi/2}+[\sin(\theta)]^{0}_{3\pi/2}$$

$$=(0-(\frac{3\pi}{2}))+(\sin(0)-\sin(\frac{3\pi}{2}))$$

$$=-\frac{3\pi-2}{2}$$

Since the intervals of the angles $\frac{-\pi}{2}$ to $0$ is the same as $\frac{3\pi}{2}$ to $0$, then why are the definite integral values different?

Also, shouldn't the value obtained by the integral in between the intervals $-\frac{\pi}{2}$ to $0$ be negative? Because the interval lies in the 4th quadrant - meaning the area under the curve lies below the x-axis and thus should have negative area.

Is my reasoning wrong? Could someone please clear these doubts of mine?

Answer 1

$$\int_{3\pi/2}^0 [1+\cos(\theta)]]^2 d\theta=-\frac{3\pi-2}{2}$$

$$\int^{3\pi/2}_0 [1+\cos(\theta)]^2 d\theta=+\frac{3\pi-2}{2}$$

because when the upper and lower limits of the integral are interchanged then the evaluated definite integral value changes sign. This is because

$$\int_a^b f(x) dx =F(b)-F(a)\text{ and }\int_b^a f(x) dx =F(a)-F(b)$$

Also the two areas of cardioid between $ \theta$ limits [red $(0, 3\pi/2)$, blue $(3\pi/2, 4 \pi/2)$] are not equal.

Answer 2

Firstly, area is never negative.

Secondly, $\displaystyle\int_{x=x_1}^{x=x_2} y(x) dx$ gives a negative value if and only if the total area below the $x-$axis is greater than the total area above the $x-$axis. In fact, the above integral is precisely equal to the area above the $x-$axis minus the area below the $x-$axis. But the point is that this is about $\int y(x) dx,$ not area or polar coordinates.

In polar coordinates, when finding area you don't usually need to think about "above" and "below" $x-$axis, because you are finding area, which is always positive anyway.

Thirdly, in polar coordinates, the area of a region bounded by two lines $\theta=\theta_1$ and $\theta=\theta_2$ and a curve $r=f(\theta),\ $ is $\quad\frac{1}{2} \displaystyle\int_{\theta=\theta_1}^{\theta=\theta_2} r^2 d\theta,\ $ and not $\quad\displaystyle\int_{\theta=\theta_1}^{\theta=\theta_2} r d\theta.$ If you apply this to your question you will find that $\quad\frac{1}{2} \displaystyle\int_{\theta=-\frac{\pi}{2}}^{\theta=0} (1+\cos(\theta))^2 d\theta = \quad\frac{1}{2} \displaystyle\int_{\theta=\frac{3\pi}{2}}^{\theta=2\pi} (1+\cos(\theta))^2 d\theta. $