Question

Principle 2 (Pappus' Theorem): Suppose \(\mathcal{S}\) has rotational symmetry about and axis Q. Find the area A of a cross section of \(\mathcal{S}\) by a plane containing the axis Q. Determine the distance R from the axis Q to the centroid of the cross section (the centroid corresponds to the center of mass if cross section were made of uniform density). Then VOLUME(\(\mathcal{S}\)) = (2\pi R) A. In other words, multiply the circumference of circle traced by the centroid as it revolves around Q with the area of the cross section. Example: Let \(\mathcal{S}\) be the torus illustrated to the right. It's cross section has area A = \(\pi r^2\). The centroid is the center of the circular cross sections, so its distance from the axis of rotation is R. Therefore VOLUME = (2\pi R)\(\pi r^2\) = \(2\pi^2 Rr^2\)

          Principle 2 (Pappus' Theorem): Suppose \(\mathcal{S}\) has rotational symmetry about and axis Q.
Find the area A of a cross section of \(\mathcal{S}\) by a plane containing the axis Q. Determine the
distance R from the axis Q to the centroid of the cross section (the centroid corresponds
to the center of mass if cross section were made of uniform density).
Then VOLUME(\(\mathcal{S}\)) = (2\pi R) A. In other words, multiply the circumference of circle
traced by the centroid as it revolves around Q with the area of the cross section.
Example: Let \(\mathcal{S}\) be the torus illustrated to the right.
It's cross section has area A = \(\pi r^2\). The centroid is
the center of the circular cross sections,
so its distance from the axis of rotation is R.
Therefore VOLUME = (2\pi R)\(\pi r^2\) = \(2\pi^2 Rr^2\)

Principle 2 (Pappus' Theorem): Suppose 𝒮 has rotational symmetry about and axis Q.
Find the area A of a cross section of 𝒮 by a plane containing the axis Q. Determine the
distance R from the axis Q to the centroid of the cross section (the centroid corresponds
to the center of mass if cross section were made of uniform density).
Then VOLUME(𝒮) = (2πR) A. In other words, multiply the circumference of circle
traced by the centroid as it revolves around Q with the area of the cross section.
Example: Let 𝒮 be the torus illustrated to the right.
It's cross section has area A = π r^2. The centroid is
the center of the circular cross sections,
so its distance from the axis of rotation is R.
Therefore VOLUME = (2πR)π r^2 = 2π^2 Rr^2

Added by Kimberly B.

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