Originally Posted by

**Chris L T521** Not quite. For the second part, you would need to do $\displaystyle 2\pi\bar y {\iint\limits_R {dA}}$

Let me simplify the first part...

We can rewrite the equation $\displaystyle \bar x = \frac{{\iint\limits_R {x\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar x = \frac{1}

{\text{Area of R}}{\iint\limits_R {x\,dA}}$ and $\displaystyle \bar y = \frac{{\iint\limits_R {y\,dA}}}{{\iint\limits_R {dA}}}$ as $\displaystyle \bar y = \frac{1}

{\text{Area of R}}{\iint\limits_R {y\,dA}}$

We can determine $\displaystyle \bar x$ two ways: by inspection or by evaluating the integral.

Since the semi-circle is symmetric about the y axis, we can determine that $\displaystyle \bar x=0$. Now if you didn't see that, then we would evaluate the equation for $\displaystyle \bar x$:

[Let me denote the radius of the circle as $\displaystyle r_0$ instead of $\displaystyle r$]...

$\displaystyle \bar x = \frac{1}

{\text{Area of R}}{\iint\limits_R {x\,dA}}=\frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta$

Note how we are using polar coordinates to evaluate the integral.

We now see that:

$\displaystyle \begin{aligned}\bar x &= \frac{1}{\frac{1}{2}\pi {r_0}^2}\int_0^{\pi}\int_0^{r_0}(r\cos\theta)r\,dr \,d\theta \\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\cos\theta\right]_{r=0}^{r_0} \\

&= \frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3}{r _0}^3\right)\int_0^{\pi}\cos\theta \,d\theta \\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\cdot 0 \\

&=\color{red}\boxed{0}\end{aligned}$

Are you still with me?

Now let's find $\displaystyle \bar y$:

$\displaystyle \begin{aligned}

\bar y &=\frac{1}{\frac{1}{2}\pi{r_0}^2}{\iint\limits_ R {y\,dA}}\\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\int_ 0^{r_0}(r\sin\theta)r\,dr\,d\theta\\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\int_0^{\pi}\l eft[\frac{1}{3}r^3\sin\theta\right]_{r=0}^{r_0}\,d\theta\\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3} {r_0}^3\right)\int_0^{\pi}\sin\theta\,d\theta\end{ aligned}$

$\displaystyle \begin{aligned} \bar y&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{1}{3 }{r_0}^3\right)\left[-\cos\theta\right]_0^{\pi}\\

&=\frac{1}{\frac{1}{2}\pi{r_0}^2}\left(\frac{2}{3} {r_0}^3\right)\\

&=\color{red}\boxed{\frac{4r_0}{3\pi}}\end{aligned }$

Thus, the centroid is $\displaystyle \color{red}\boxed{\bigg(0,\frac{4r_0}{3\pi}\bigg)}$

This answers your first question.

Now the second:

Using the theorem of Pappus, we can find this volume:

$\displaystyle V=2\pi \cdot \bar y \cdot \left(\text{Area of R}\right)$

$\displaystyle \therefore V=2\pi\cdot \left[\frac{4r_0}{3\pi}\right]\cdot \left[\frac{1}{2}\pi {r_0}^2\right]=\color{red}\boxed{\frac{4}{3}\pi{r_0}^3}$

We know this to be true because the volume of a sphere is $\displaystyle \frac{4}{3}\pi r^3$

I hope you can absorb all of this. If you have questions, feel free to ask!

--Chris