Question

Someone please help me to find the centroid of a semi circle using polar coordinates and a double integral? With a radius of r, and theta being between 0 and pi

Answer 1

The area for a semi circle in Cartesian coordinate system with the flat edge against the x-axis is:
\[A = \int\limits_{0}^{\sqrt{r^{2}-x^{2}}}\int\limits_{-r}^{r}dxdy\]
The bounding limits for x are from -r to r. And for y, this comes from the equation for a circle:
\[x ^{2}+y ^{2}=r ^{2}\]
Solve for y:
\[y = \pm \sqrt{r ^{2} - x ^{2}}\]
But since we are only dealing with a semi circle, we can look at the positive answer. (Either or would give you the same result).

Now to convert to polar coordinate system, we use some equations you have probably seen:
\[x = rcos \theta\]\[y = rsin \theta\]\[dxdy = rdrd \theta\]
Substitute to convert to polar coordinate system:
\[A = \int\limits_{0}^{\pi}\int\limits_{0}^{r}rdrd \theta\]Solve this for your area.
\[A = \pi r ^{2}/2\](Let me know if you are having issues with integrating double integrals)

To find your x coordinate of the centroid:
\[x = 1/A \int\limits_{}^{}\int\limits_{}^{}xdxdy\]or\[x = 1/A \int\limits_{0}^{\pi}\int\limits_{0}^{r}rcos \theta (rdrd \theta)\]Solve this and you will find x = 0 which intuitively makes sense.

Similar for the y coordinate of the centroid:
\[y = 1/A \int\limits_{}^{}\int\limits_{}^{}ydxdy\]or\[y = 1/A \int\limits_{0}^{\pi}\int\limits_{0}^{r}rsin \theta (rdrd \theta)\]Solve this and you will find \[y = 4r/3\pi\]
This means the centroid of a semicircle is at: \[(0,4r/3\pi)\]

Sorry for the long draw out solution but it's not easy to explain double integrals in text. If you have ANY questions or issues, please post them here so we all can learn.