C1: $\;r= 3 \sin x$ where $0\leq x\leq \pi$
C2: $\;r= 1 + \sin x$ where $-\pi \leq x \leq \pi$
Please help me in this question
I have drawn the sketch of the two polar equations. Then the question says - Show that the area of the region which is inside C1 but outside C2 is π.
^^^ this statement I believe it's the area that I shaded in red, BUT in the marking scheme the examiner has done it as C2 total area - C1 total area. But in the diagram it is obvious that the curve C2 is not fully inside C1 for this to be true, there is that small part outside , so I have no idea how they subtracted area of whole C2.
I am going to write $x=\theta$ to emphasise that $\theta$ is an angle and not a horizontal coordinate. You need to compute the polar coordinates of the points where $C_1$ and $C_2$ meet, by solving $3sin(\theta)=1+sin(\theta)$. You should get $\theta=\pi/6$ or $\theta=5\pi/6$, and $r=3/2$. Then just compute the area of each curve between these values of $\theta$, and subtract.
Since we are only considering the curves between these values of $\theta$, the issue with 'one part of $C_2$ being outside $C_1$' does not arise.
The area which you have shaded is correct.