Satellite Dish - Max Amount and Tilt

TheJason

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What is the max amount of water the dish will hold, expressed as a triple integral, for parabolic dish that is \(\displaystyle 3\, m\) wide and \(\displaystyle \dfrac{1}{2}\, m\) deep? Consider the fact that it's axis of symmetry has a tilt \(\displaystyle 25\) degrees from the vertical.

Second, what would be the smallest tilt of the dish where it can contain no water?
 

MarkFL

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I think the first thing I would do is find the equation of the paraboloid. A circular paraboloid with its vertex at the origin has the form:

\(\displaystyle z=a\left(x^2+y^2\right)\)

Now, when \(\displaystyle z=\frac{1}{2}\) we require \(\displaystyle x^2+y^2=\left(\frac{3}{2}\right)^2\) and this implies:

\(\displaystyle a=\frac{2}{9}\)

And so our paraboloid is:

\(\displaystyle z=\frac{2}{9}\left(x^2+y^2\right)\)

The surface of the water will be a plane...how can we find this plane?
 
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