# Question #678c2

Nov 15, 2017

The ellipsoid has surface area ${\pi}^{2} a b$.

#### Explanation:

The ellipsoid is generated by the circles drawn into the figure, which form loops around the x-axis. So, the surface area should be equal to the sum of the circumferences of each circle.

The radius of each circle is given by its height $y$. The radii of the circles range between $0$ and $b$.

We should write a function describing this. Recalling the equation for ellipses, note that the function here is ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$.

This yields the simplification: ${y}^{2} = {b}^{2} \left(1 - {x}^{2} / {a}^{2}\right) = {b}^{2} / {a}^{2} \left({a}^{2} - {x}^{2}\right)$.

The height is then given by $y = \frac{b}{a} \sqrt{{a}^{2} - {x}^{2}}$.

This is the radius of the circle, so our circumference will be given at each point by $C = 2 \pi r = 2 \pi y = \frac{2 \pi b}{a} \sqrt{{a}^{2} - {x}^{2}}$.

So, all we need to do is add up these circumferences. We can find the circles just to the right of the y-axis by integrating from $0$ to $a$ then doubling that for the surface area of the entire ellipsoid. So the surface area $S$ is:

$S = 2 {\int}_{0}^{a} \frac{2 \pi b}{a} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = \frac{4 \pi b}{a} {\int}_{0}^{a} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx}$

To perform this integration, we could do some tedious work with the substitution $x = a \sin \theta$.

A quicker method, however, would be to realize this is integral describing the area under the curve $y = \sqrt{{a}^{2} - {x}^{2}}$, which is the same as the circle centered at the origin with radius $a$, ${x}^{2} + {y}^{2} = {a}^{2}$.

We care only for the positive part (the square root is positive) and from $0$ to $a$, which is the first quadrant, or a fourth of the circle. A fourth of the area of the circle is $\frac{\pi {a}^{2}}{4}$.

Then:

$S = \frac{4 \pi b}{a} \left(\frac{\pi {a}^{2}}{4}\right) = {\pi}^{2} a b$

Nov 15, 2017

The surface area of the torus is $4 {\pi}^{2} a b$.

#### Explanation:

The torus is formed by swinging a circle of radius $b$ in a larger circle of radius $a$.

The surface area of the torus will be the collective sum of all the circumferences of the circles with radius $b$, where $C = 2 \pi b$.

Well, we're doing this along the entire distance of $2 \pi a$, which is the circumference of the circle we're sweeping the littler circle along.

So the surface area would simply be $2 \pi b \left(2 \pi a\right) = 4 {\pi}^{2} a b$.