# Thread: Using a double integral to find the area of a circle.

1. ## Using a double integral to find the area of a circle.

I have scanned the question I am working on and attached it as an image to this thread. A hint on this question would be good enough.

2. Originally Posted by Undefdisfigure
I have scanned the question I am working on and attached it as an image to this thread. A hint on this question would be good enough.
Is this a unit circle?

3. Originally Posted by Undefdisfigure
I have scanned the question I am working on and attached it as an image to this thread. A hint on this question would be good enough.
Note that $\displaystyle r=\cos\theta$ is a circle with center at $\displaystyle \left(\frac{1}{2},0)$ and radius $\displaystyle r=\frac{1}{2}$.

Similarly, $\displaystyle r=\sin\theta$ is a circle with center at $\displaystyle \left(0,\frac{1}{2})$ and radius $\displaystyle r=\frac{1}{2}$.

Now these circles will overlap at some point. So my hint is this:

When considering the region of intersection, split it up into two separate regions. Due to how we find limits of integration in polar coordinates, each region will be defined over different ranges for $\displaystyle \theta$. So your integrals will look something like:

$\displaystyle \displaystyle\int_{\theta_0}^{\theta_1} \int_0^{\cos\theta} r\,dr\,d\theta+\int_{\theta_0^{\prime}}^{\theta_1^ {\prime}}\int_0^{\sin\theta}r\,dr\,d\theta$

where $\displaystyle \theta_0\leq \theta\leq\theta_1$ is the range you consider for the $\displaystyle r=\cos\theta$ part, and $\displaystyle \theta_0^{\prime}\leq \theta\leq\theta_1^{\prime}$ is the range you consider for the $\displaystyle r=\sin\theta$ part.

Can you proceed? If you still have issues, post back!

4. Oops, I didn't see somebody else posted.

5. Originally Posted by Undefdisfigure
Yeah but what kind of limits of integration would I use?
To figure out the limits of integration for $\displaystyle \theta$, its best to look at a graph showing the intersection of the two circles. The only additional hint I'll supply for now is that $\displaystyle \theta=\frac{\pi}{4}$ is a value worth considering.

,
,

,

,

,

,

,

,

,

,

# find area of circle using double integration

Click on a term to search for related topics.