# Thread: Polar Equations and Areas

1. ## Polar Equations and Areas

I'm having a difficult time applying the formula $\frac{1}{2}\int_{\alpha}^{\beta}r^2d\theta$ for areas under curves in polar form. My problem is deducing the limits of integration. For example, if I want to find the area of the interior for $r=1-sin(\theta)$, I graph the function on my calc, and then I trace from the origin back to the y-axis. My calculator gives the angles in radians, which are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, so I can integrate $2\left(\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[1-sin(\theta)]^2d\theta\right)$ to find the area.

What method can I use, in general, to find the limits of integration for problems like this? How can I determine on paper the interval through which the curve is traced?

2. While I'm not sure to be understanding what you are doing, I feel like you are not familiar with trigonometric graphs.

y = sin(x)

Is a curve starting at 0, going up to 1 at 90 degrees, goes back to 0 at 180 degrees, going further down till -1 at 270 degrees and up again to 0 at 360 degrees.

y &#61;sin&#40;x - Wolfram|Alpha)

Since you put the negative sign in front of the sine, you reflect the graph through the x-axis, giving:

y &#61;-sin&#40;x - Wolfram|Alpha)

Now, there is 1 in front. This means that you 'raise' the curve one unit up. Zeros thus become 1, 1s become 2s and -1s becomes 0s, like this:

y &#61;-sin&#40;x&#41; &#43;1 - Wolfram|Alpha

Now, if you can sketch such graphs on paper, knowing where the maxima, minima, and where the curve cuts the axes, you'll be better off finding the limits.

I hope it helped!

3. Originally Posted by Unknown008
While I'm not sure to be understanding what you are doing, I feel like you are not familiar with trigonometric graphs.

y = sin(x)

Is a curve starting at 0, going up to 1 at 90 degrees, goes back to 0 at 180 degrees, going further down till -1 at 270 degrees and up again to 0 at 360 degrees.

y &#61;sin&#40;x - Wolfram|Alpha)

Since you put the negative sign in front of the sine, you reflect the graph through the x-axis, giving:

y &#61;-sin&#40;x - Wolfram|Alpha)

Now, there is 1 in front. This means that you 'raise' the curve one unit up. Zeros thus become 1, 1s become 2s and -1s becomes 0s, like this:

y &#61;-sin&#40;x&#41; &#43;1 - Wolfram|Alpha

Now, if you can sketch such graphs on paper, knowing where the maxima, minima, and where the curve cuts the axes, you'll be better off finding the limits.

I hope it helped!
These are polar equations. $r=1-sin(\theta)$ in a polar coordinate system (where points are represented by a radial distance from the origin and an angle $(r,\theta)$) is much different then the function $f(x)=1-sin(x)$ on the x-y plane. The graph of the polar form is a closed heart-shaped structure.

4. I'll try to explain how I do this: Imagine that you have a clock on the origin with only one pointer. The pointer direction depends on each graph you deal, but it necessarily will rest, initially, in one of the cardinal points (north, south, west, east).

So, imagine that this pointer will start rotating to some direction - think that it will sweep the area that you want to find. You "stop" either when the graph "ends" (in other words, when you swept the region you wanted) or when it completes a full turn (360 deg or $2\pi$ rad).

If it completes a full turn, then the limits will be from 0 to $2\pi$. In the cardioid $r=1-\sin(\theta)$ you do a full turn, while, for example, if you look the polar function $r = \sin(\theta)$ in the figure, the limits of the area integral will be from 0 to $\pi$, if you start from east to west.

On the other hand, the limits of the function $r=\cos(\theta)$ can be found by starting the "pointer" south and then going north:

In other words, you go from $-\pi/2$ to $\pi/2$.

I hope you understand this tip hehe! Good luck.