First Pappus-Guldinus Theorem

Theorem
Let $C$ be a plane figure that lies entirely on one side of a straight line $\LL$.

Let $S$ be the solid of revolution generated by $C$ around $\LL$.

Then the volume of $S$ is equal to the area of $C$ multiplied by the distance travelled by the centroid of $C$ around $\LL$ when generating $S$.

Proof
Let $V$ denote the volume of $S$

Let $\AA$ denote the area of $C$.

Let $C$ be embedded in a Cartesian plane such that $\LL$ coincides with the $x$-axis.

Let $\tuple {\overline x, \overline y}$ be the coordinates of the centroid of $C$.

Consider a rectangle $R$ bounded by the lines:
 * $y = 0$
 * $x = \xi$
 * $x = \xi + \delta x$
 * $y = \map f x$

The moment $M_y$ of $R$ about the $y$-axis is given by:
 * $M_y = \map f x \xi \rdelta \xi$

Hence from Area under Curve:
 * $\AA \overline x = \ds \int_a^b x \map f x \rd x$

The moment $M_x$ of $R$ about the $x$-axis is given by:
 * $M_x = y \rdelta x \dfrac y 2$

that is, half way up.

Hence:
 * $\AA \overline y = \dfrac 1 2 \ds \int_a^b y^2 \rd x$

It follows immediately that:
 * $\dfrac V \AA \overline y = 2 \pi$

That is:
 * $V = 2 \pi \AA \overline y$

Also known as
This result is also known as:
 * Pappus's Centroid Theorem for Volume
 * the First Guldinus Theorem.

Also see

 * Second Pappus-Guldinus Theorem