Question #678c2

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Question #678c2

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Answer 1 · 2017-11-15T23:49:16

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

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 $y$ $y$ . The radii of the circles range between $0$ $0$ and $b$ $b$ .

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

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

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

This is the radius of the circle, so our circumference will be given at each point by $C = 2 π r = 2 π y = \frac{2 π b}{a} \sqrt{a^{2} - x^{2}}$ $C=2pir=2piy=(2pib)/asqrt(a^2-x^2)$ .

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 $0$ $0$ to $a$ $a$ then doubling that for the surface area of the entire ellipsoid. So the surface area $S$ $S$ is:

$S = 2 \int_{0}^{a} \frac{2 π b}{a} \sqrt{a^{2} - x^{2}} d x = \frac{4 π b}{a} \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x$ $S=2int_0^a(2pib)/asqrt(a^2-x^2)dx=(4pib)/aint_0^asqrt(a^2-x^2)dx$

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

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

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

Then:

$S = \frac{4 π b}{a} (\frac{π a^{2}}{4}) = π^{2} a b$ $S=(4pib)/a((pia^2)/4)=pi^2ab$

Answer 2 · 2017-11-15T23:54:54

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

Explanation:

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

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

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

So the surface area would simply be $2 π b (2 π a) = 4 π^{2} a b$ $2pib(2pia)=4pi^2ab$ .

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