Volume of a ring

The volume of the elliptical torus can be found somewhat easily using Pappus' Centroid Theorems . The area of an ellipse is $A=\pi ab$ , where $a$ and $b$ are the semi-axes of the ellipse. The area of the ellipse in this problem is ${\large\frac{21}{16}}\pi$ mm $^2$ . The centroid of the ellipse is $10\text{ mm}$ away from the axis of rotation, and so the centroid passes a distance of $20\pi\text{ mm}$ through its rotation.

By Pappus, the volume of revolution is ${\large\frac{21}{16}}\pi\times20\pi={\large\frac{105}{4}}\pi^2\approx\boxed{259.077}\text{ mm}^3$

Now for something a bit more challenging... doing the problem with the Shell Method :

Let the $y$ axis be the axis of rotation. The formula for the ellipse is: $\frac{16(x-10)^2}{9}+\frac{16y^2}{49}=1$

Solving for $y$ yields: $y=\pm\frac{7}{4}\sqrt{1-\frac{16}{9}(x-10)^2}$

Using the shell method, the integral to obtain the volume will be: $V=7\pi\int_{9.25}^{10.75} x \sqrt{1-\frac{16}{9}(x-10)^2} \, dx$

Note that we had to multiply the integral by $2$ because the expression for $y$ only gives us the top half of the ellipse.

This integral can be found using Trigonometric Substitution .

$V=7\pi\left[-\frac{3}{16}\left(1-\frac{16}{9}(x-10)^2\right)^{3/2}+5(x-10)\sqrt{1-\frac{16}{9}(x-10)^2}+\frac{15}{4}\arcsin\left(\frac{4}{3}(x-10)\right)\right]_{9.25}^{10.75}$

This evaluates to ${\large\frac{105}{4}}\pi^2\approx\boxed{259.077}\text{ mm}^3$

volume of revolution = $2 \pi R A = 2 \pi (10)(\pi \times \frac{3.5}{2} \times \frac{1.5}{2}) = \textbf{259.077}$

This ring shape is a torus that has been stretch vertically by a scale of $\frac{3.5mm}{1.5mm}$ (Think of it as if we were to squash this shape down so the major axis of the elliptical cross section equals 1.5mm, we would get a torus). The volume of a torus is given by $2\pi^2r^2R$ where $r$ is the diameter of the circle cross-section and $R$ is the distance of the centre of the torus to the center of the circle cross-section. In this case, $r=0.75mm$ and $R=10mm$ which gives us a volume of $111.03305mm^3$ . Multiply this by $\frac{3.5mm}{1.5mm}$ which gives an answer of $259.077mm^3$