1
$\begingroup$

Find the area of the region $R$ given by two curves.

So the region $R$ describes the area that is common between the two curves: $$\begin{align*} \text{Function 1: } r&= 2\sin(\theta)\\ \text{Function 2: }r&= \frac{3}{2} - \sin(\theta). \end{align*}$$

I have to find the area that is common between the two graphs..

What I did was just take the integral of function 1 (from 0 to $2\pi$) then subtract it from the integral of function 2 (from 0 to $2\pi$)... which gave me ($2-3\pi$) as my final answer... is this correct?

  • 1
    It is not correct, unless it so happens that the second graph is completely contained inside the first graph (which it is not). Try graphing the curves, and find the points of intersection.2012-06-10
  • 0
    I did graph them... they intersect at (1,1) and (-1,1). For the sin graph, I flipped at about the x axis as well, so there are two circles centered at (0,1) and (0,-1)2012-06-10
  • 0
    Which correspond to certain values of $\theta$. So you want to do the integral that "sweeps" **only** the area of intersection. That is most certainly not the integral from $\theta=0$ to $\theta=2\pi$.2012-06-10
  • 0
    But I have overlapping occuring in all 4 quadrants... wouldnt that mean I have to account for the full 2pi?2012-06-10
  • 1
    The answer you got cannot possibly be correct as it is a negative number...2012-06-10
  • 0
    @Nick: No, because that includes regions where we do **not** have area enclosed in both graphs. You need to think about it a lot more carefully than you are.2012-06-10
  • 0
    @Nick: $r=2\sin\theta$ is a circle of radius $1$ centered at $(0,1)$. It does not even *go* through quadrants III and IV, so how could you possibly have overlap "in all 4 quadrants"?2012-06-10
  • 0
    Per chance, after careful analysis of the regions, one could also avail of the change of variables?2012-06-10

1 Answers 1