I am trying to calculate the centroid of the solid of revolution defined by $y=\sin (x)$ from $x=0$ to $x=\pi$ rotated around the $x$ axis. All the information I've been able to find online relates either to centroids of surface areas, or centroids of 2d areas. What is the formula for me to use here?
[Math] Centroid of a solid of revolution
calculuscentroidsolid of revolution
Related Solutions
Let a shape $S$ in the $(x,y)$-plane be defined by
$$a(x)\leq y\leq b(x)\qquad(0\leq x\leq R)\ .$$
When $S$ is rotated around the $y$-axis we obtain a rotational body $B$. This body can be seen as a union of thin cylindrical shells of radius $x$, height $b(x)-a(x)$, and thickness $dx$. The volume of such a shell is $dV=2\pi x\bigl(b(x)-a(x)\bigr)\ dx$, and its centroid is on the $y$-axis at level $h(x):={1\over2}\bigl(a(x)+b(x)\bigr)$.
The level $\eta$ of the centroid of the full body $B$ is the weighted mean of the levels $h(x)$ of these shells. Therefore
$$\eta\ =\ {\int\nolimits _0^R h(x)\ dV \over \int\nolimits_0^R dV}\ = {\int\nolimits _0^R x\bigl(b^2(x)-a^2(x)\bigr)\ dx \over 2 \int\nolimits_0^R x\bigl(b(x)-a(x)\bigr)\ dx}\ .$$
Of course it is easily possible to translate this "stenographic" derivation into a fullfledged proof by partitioning the $x$-interval $[0,R]$ into $N$ equal parts and letting $N\to\infty$.
$\newcommand{\dd}{\partial}$If $\bigl(x(t), z(t)\bigr)$, $a \leq t \leq b$, parametrizes a smooth plane curve $C$ in the half-plane $x > 0$, the surface $S$ swept out by revolving $C$ about the $z$-axis may be parametrized by $$ X(s, t) = \bigl(x(t) \cos s, x(t) \sin s, z(t)\bigr),\qquad a \leq t \leq b,\quad 0 \leq s \leq 2\pi. $$ The partial derivatives are \begin{align*} \frac{\dd X}{\dd s} &= \bigl(-x(t) \sin s, x(t) \cos s, 0\bigr), \\ \frac{\dd X}{\dd t} &= \bigl(x'(t) \cos s, x'(t) \sin s, z'(t)\bigr); \end{align*} Their cross product is $$ \frac{\dd X}{\dd s} \times \frac{\dd X}{\dd t} = -x(t) \bigl(z'(t) \cos s, z'(t) \sin s, x'(t)\bigr); $$ the area element is $$ \left\lVert\frac{\dd X}{\dd s} \times \frac{\dd X}{\dd t}\right\rVert ds\, dt = x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, ds\, dt. \tag{1} $$ The surface area of $S$ is $$ \int_{a}^{b} \int_{0}^{2\pi} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, ds\, dt = 2\pi \int_{a}^{b} x(t)\sqrt{z'(t)^{2} + x'(t)^{2}}\, dt. $$ If $$ \ell = \int_{a}^{b} \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt $$ denotes the arc length of $C$, the area of $S$ becomes $$ 2\pi \int_{a}^{b} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt = 2\pi\, \ell \left(\frac{1}{\ell} \int_{a}^{b} x(t) \sqrt{z'(t)^{2} + x'(t)^{2}}\, dt\right) = \ell\, (2\pi\, \bar{x}), $$ the length of $C$ times the circumference of the circle swept by the centroid of $C$.
Best Answer
An alternative method is to use Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area A and the length of the path traced by its centroid R, i.e., 2πR. When composite areas are involved, the centroid is the weighted sum of the component centroids. The bottom line is that the volume is given simply by V=2πRA.
For purpose of symmetry, I redefine your problem by shifting the axis so that $y=\text{cos}(x)$ from $x=-\pi/2$ to $x=\pi/2$. Now, without any further ado, the centroid is at $R=\pi/8$ and the area under the curve is $A=2$. Then
$$V=2\pi RA=2\pi (\pi/8)2=\pi ^2/2 $$
Now granted, you need to calculate the centroid. And that's why I shifted the axis. Not many people know that for a symmetric function such as this the centroid can be expressed as the area of square of the curve, divided by the twice the are under the curve, i.e.,
$$R=\frac{\int_{-\pi/2}^{\pi/2}\text{cos}^2 (x) dx}{2\int_{-\pi/2}^{\pi/2}\text{cos} (x) dx} $$
The point of this exercise is that you have a more general rule that can be applied to many problems, and not just an ad hoc solution to a single one.