Proof

A curve given by the positive function ${\displaystyle f(x)}$ is bounded by two points given by:

${\displaystyle a\geq 0}$ and ${\displaystyle b\geq a}$

If ${\displaystyle dL}$ is an infinitesimal line element tangent to the curve, the length of the curve is given by:

$${\displaystyle L=\int _{a}^{b}dL=\int _{a}^{b}{\sqrt {dx^{2}+dy^{2}}}=\int _{a}^{b}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx}$$

The ${\displaystyle y}$ component of the centroid of this curve is:

$${\displaystyle {\bar {y}}={\frac {1}{L}}\int _{a}^{b}y\,dL={\frac {1}{L}}\int _{a}^{b}y{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx}$$

The area of the surface generated by rotating the curve around the x-axis is given by:

$${\displaystyle A=2\pi \int _{a}^{b}y\,dL=2\pi \int _{a}^{b}y{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx}$$

Using the last two equations to eliminate the integral we have:

$${\displaystyle A=2\pi {\bar {y}}L}$$

Proof 1

The area bounded by the two functions:

$${\displaystyle y=f(x),\,\qquad y\geq 0}$$

$${\displaystyle y=g(x),\,\qquad f(x)\geq g(x)}$$

and bounded by the two lines:

${\displaystyle x=a\geq 0}$ and ${\displaystyle x=b\geq a}$

is given by:

$${\displaystyle A=\int _{a}^{b}dA=\int _{a}^{b}[f(x)-g(x)]\,dx}$$

The ${\displaystyle x}$ component of the centroid of this area is given by:

$${\displaystyle {\bar {x}}={\frac {1}{A}}\,\int _{a}^{b}x\,[f(x)-g(x)]\,dx}$$

If this area is rotated about the y-axis, the volume generated can be calculated using the shell method. It is given by:

$${\displaystyle V=2\pi \int _{a}^{b}x\,[f(x)-g(x)]\,dx}$$

Using the last two equations to eliminate the integral we have:

$${\displaystyle V=2\pi {\bar {x}}A}$$

Proof 2

Let ${\displaystyle A}$ be the area of ${\displaystyle F}$, ${\displaystyle W}$ the solid of revolution of ${\displaystyle F}$, and ${\displaystyle V}$ the volume of ${\displaystyle W}$. Suppose ${\displaystyle F}$ starts in the ${\displaystyle xz}$-plane and rotates around the ${\displaystyle z}$-axis. The distance of the centroid of ${\displaystyle F}$ from the ${\displaystyle z}$-axis is its ${\displaystyle x}$-coordinate

$${\displaystyle R={\frac {\int _{F}x\,dA}{A}},}$$

and the theorem states that

$${\displaystyle V=Ad=A\cdot 2\pi R=2\pi \int _{F}x\,dA.}$$

To show this, let ${\displaystyle F}$ be in the xz-plane, parametrized by ${\displaystyle \mathbf {\Phi } (u,v)=(x(u,v),0,z(u,v))}$ for ${\displaystyle (u,v)\in F^{*}}$, a parameter region. Since ${\displaystyle {\boldsymbol {\Phi }}}$ is essentially a mapping from ${\displaystyle \mathbb {R} ^{2}}$ to ${\displaystyle \mathbb {R} ^{2}}$, the area of ${\displaystyle F}$ is given by the change of variables formula:

$${\displaystyle A=\int _{F}dA=\iint _{F^{*}}\left|{\frac {\partial (x,z)}{\partial (u,v)}}\right|\,du\,dv=\iint _{F^{*}}\left|{\frac {\partial x}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial x}{\partial v}}{\frac {\partial z}{\partial u}}\right|\,du\,dv,}$$

where ${\displaystyle \left|{\tfrac {\partial (x,z)}{\partial (u,v)}}\right|}$ is the determinant of the Jacobian matrix of the change of variables.

The solid ${\displaystyle W}$ has the toroidal parametrization ${\displaystyle {\boldsymbol {\Phi }}(u,v,\theta )=(x(u,v)\cos \theta ,x(u,v)\sin \theta ,z(u,v))}$ for ${\displaystyle (u,v,\theta )}$ in the parameter region ${\displaystyle W^{*}=F^{*}\times [0,2\pi ]}$; and its volume is

$${\displaystyle V=\int _{W}dV=\iiint _{W^{*}}\left|{\frac {\partial (x,y,z)}{\partial (u,v,\theta )}}\right|\,du\,dv\,d\theta .}$$

Expanding,

$${\displaystyle {\begin{aligned}\left|{\frac {\partial (x,y,z)}{\partial (u,v,\theta )}}\right|&=\left|\det {\begin{bmatrix}{\frac {\partial x}{\partial u}}\cos \theta &{\frac {\partial x}{\partial v}}\cos \theta &-x\sin \theta \\[6pt]{\frac {\partial x}{\partial u}}\sin \theta &{\frac {\partial x}{\partial v}}\sin \theta &x\cos \theta \\[6pt]{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&0\end{bmatrix}}\right|\\[5pt]&=\left|-{\frac {\partial z}{\partial v}}{\frac {\partial x}{\partial u}}\,x+{\frac {\partial z}{\partial u}}{\frac {\partial x}{\partial v}}\,x\right|=\ \left|-x\,{\frac {\partial (x,z)}{\partial (u,v)}}\right|=x\left|{\frac {\partial (x,z)}{\partial (u,v)}}\right|.\end{aligned}}}$$

The last equality holds because the axis of rotation must be external to ${\displaystyle F}$, meaning ${\displaystyle x\geq 0}$. Now,

$${\displaystyle {\begin{aligned}V&=\iiint _{W^{*}}\left|{\frac {\partial (x,y,z)}{\partial (u,v,\theta )}}\right|\,du\,dv\,d\theta \\[1ex]&=\int _{0}^{2\pi }\!\!\!\!\iint _{F^{*}}x(u,v)\left|{\frac {\partial (x,z)}{\partial (u,v)}}\right|du\,dv\,d\theta \\[6pt]&=2\pi \iint _{F^{*}}x(u,v)\left|{\frac {\partial (x,z)}{\partial (u,v)}}\right|\,du\,dv\\[1ex]&=2\pi \int _{F}x\,dA\end{aligned}}}$$

by change of variables.