Chocolate Covered Candy

Phil the confectioner is making a new batch of chocolate covered candy. Each candy centre is shaped like an ellipsoid of revolution defined by the equation:
$b^2 x^2 + b^2 y^2 + a^2 z^2 = a^2 b^2$.
Phil wants to know how much chocolate is needed to cover one candy centre with a uniform coat of chocolate one millimeter thick.

If $a = 1$ mm and $b = 1$ mm, the amount of chocolate required is $\dfrac{28}{3} \pi$ mm3
If $a = 2$ mm and $b = 1$ mm, the amount of chocolate required is approximately 60.35475635 mm3.
Find the amount of chocolate in mm3 required if $a = 3$ mm and $b =1$ mm. Give your answer as the number rounded to 8 decimal places behind the decimal point.

We know that the volumes of two ellipsoids are related according to the change in their semi-axes lengths. So, if we scale one axis of the ellipsoid (without rotation), its volume will scale automatically. As a result, the volume of an ellipsoid of revolution defined by the equation $b^2 x^2 + b^2 y^2 + a^2 z^2 = a^2 b^2$ is given by $\frac{4}{3} \pi a b^2$.

For the candy, the chocolate shell is formed by the difference between the volume of the outer ellipsoid (with semi-axes length of $a+1$ mm and $b+1$ mm) and the inner ellipsoid (with semi-axes of a and b). Using the formula above, our calculation for the volume of chocolate used becomes:

$V_{chocolate} = \frac{4}{3} \pi (a+1) (b+1)^2 – \frac{4}{3} \pi a b^2$.

We can simplify this equation by factoring out $\frac{4}{3} \pi$ to get:

$V_{chocolate} = \frac{4}{3} \pi \left[(a+1) (b+1)^2 – a b^2\right]$.

Let’s plug in the values $a = 3$ mm and $b = 1$ mm to get:

$V_{chocolate} = \frac{4}{3} \pi \left[(3+1) (1+1)^2 – 3 \cdot 1^2\right]$.

Simplifying further,

$V_{chocolate} = \frac{4}{3} \pi \left[4 \cdot 2^2 – 3\right] = \frac{4}{3} \pi [16 – 3] = \frac{4}{3} \pi \cdot 13$.

This gives us

$V_{chocolate} \approx 68.7528216 \ mm^3$,

rounded to 8 decimal places.

So, if $a = 3$ mm and $b =1$ mm, the amount of chocolate required is approximately 68.7528216 mm^3.

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