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## Homework Statement

[/B]

The diameter and depth of an antenna that is

shaped like a paraboloid are ##d = 2.0m## and ##s = 0.5m## respectively. The antenna is set up so that its axis of symmetry is at an angle ##\theta = 30^{\circ}## from it's usual vertical orientation. How much can the antenna gather water in this position?

Hint: set up your coordinate axes so that the axis of symmetry of the paraboloid points in the z-direction and its vertex is at the origin. In this case you can assume that the equation of the paraboloid is of the form ##z=a(x^2 + y^2)## and that the surface of the water is going to settle on a plane ##z=bx+c##. Present the volume of the water as an integral in the xyz-coordinates,

**but do not try to calculate it by hand.**Use Matlab to approximate the value with the command ##vpa(I)##, where ##I## is the asked integral.

Correct answer: ##25 \ liters \ = 25dm^3 = 25 \times 10^{-3}m^3##

## Homework Equations

This is going to turn into the evaluation of either a two- or a three-dimensional integral, so the volume is either given by:

\begin{equation}

V = \iint_R f(x,y) dA

\end{equation}

or

\begin{equation}

V = \iiint_T 1 \ dV

\end{equation}

## The Attempt at a Solution

Since the antenna is at ##30^{\circ}## from the vertical position, if we set the coordinate axes as instructed, the slope ##b## of the plane delimiting the volume of water from above ##z = bx + c## equals ##tan\theta = tan(30^{\circ}) = \frac{1}{\sqrt{3}}##.

Since the plane touches the edge of the antenna dish, looking at the diameter and depth of the antenna gives us ##z = 1/2## when ##x = 1 \implies 1/2 = \frac{1}{\sqrt{3}} (1) + c \iff c = 1/2 - 1/\sqrt{3}##.

Therefore the equation of the plane becomes

\begin{equation}

z = \frac{1}{\sqrt{3}}x + \frac{1}{2} - \frac{1}{\sqrt{3}}

\end{equation}

Below this plane the volume is delimited by the paraboloid surface ##z = a(x^2 + y^2)##, where ##a(x^2 + y^2)## is the square of the radius ##r## of the circle projected onto the xy-plane by the paraboloid. Now because of its rotational symmetry, when ##x = 1## and ##y=1##, again ##z=1/2##. Then since ##d = 2 \implies r=1## and

[tex]

a(x^2 + y^2) = a(2) = 1 \iff a = \frac{1}{2}

[/tex]

Therefore the equation of the paraboloid is of the form

\begin{equation}

z = \frac{1}{2}(x^2 + y^2)

\end{equation}

Now to find out where these two surfaces intersect (to find out the projection of the volume onto the xy-plane) we set these two surfaces to be equal. Then

\begin{align*}

\frac{1}{\sqrt{3}}x + \frac{1}{2} - \frac{1}{\sqrt{3}} &= \frac{1}{2}(x^2 + y^2) \\

\iff \frac{2}{\sqrt{3}}x + 1 - \frac{2}{\sqrt{3}} &= (x^2 + y^2)\\

\iff x^2 - \frac{2}{\sqrt{3}}x + y^2 &= 1-\frac{2}{\sqrt{3}} || \text{ Complete the square: } + \frac{1}{3}\\

\iff x^2 - \frac{2}{\sqrt{3}}x + \frac{1}{3} + y^2 &= \frac{4}{3}-\frac{2}{\sqrt{3}}\\

\iff (x-\frac{1}{\sqrt{3}})^2 + y^2 &= \frac{4}{3}-\frac{2}{\sqrt{3}}\\

\end{align*}

This is a circle centered at ##(\frac{1}{\sqrt{3}},0)## whose radius is ##r_c = \sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}} \approx 0.42##.

Now to set up the integral. We need to solve for x in the equation of the circle if we want to establish the limits of integration:

\begin{equation*}

x = \pm \sqrt{\frac{4}{3} - \frac{2}{\sqrt{3}} - y^2} +\frac{1}{\sqrt{3}}

\end{equation*}

Therefore when ##y## goes from ##-\sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}## to ##\sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}##, ##x## goes from ##-\sqrt{\frac{4}{3} - \frac{2}{\sqrt{3}} - y^2} +\frac{1}{\sqrt{3}}## to ##\sqrt{\frac{4}{3} - \frac{2}{\sqrt{3}} - y^2} +\frac{1}{\sqrt{3}}##. Finally we integrate in the ##z##-direction, where z goes from ##\frac{1}{2}(x^2 + y^2)## to ##\frac{1}{\sqrt{3}}x + \frac{1}{2} - \frac{1}{\sqrt{3}}##.

This gives us the triple integral (f me...)

\begin{align*}

V

&=\int_{-\sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}}^{\sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}}

\int_{-\sqrt{\frac{4}{3} - \frac{2}{\sqrt{3}} - y^2} +\frac{1}{\sqrt{3}}}^{\sqrt{\frac{4}{3} - \frac{2}{\sqrt{3}} - y^2} +\frac{1}{\sqrt{3}}}

\int_{\frac{1}{2}(x^2 + y^2)}^{\frac{1}{\sqrt{3}}x + \frac{1}{2} - \frac{1}{\sqrt{3}}} 1 \ dzdxdy

\end{align*}

I am ##not## doing this by hand. Using the code below I get a complex answer:

Code:

```
%Calculating the volume of a puddle collected by a tilted paraboloid
%antenna
syms x y theta r z
f(x,y) = x/sqrt(3) + 1/2 - 1/sqrt(3);
g(x,y) = 1/2 * (x^2 + y^2);
format rat
%Innermost integral
a = int(1,z,1/2 * (x^2 + y^2),1/sqrt(3) + 1/2 - 1/sqrt(3));
%Second innermost integral
b = int(a,x,-sqrt(4/3 - 2/sqrt(3) - y^2),sqrt(4/3 - 2/sqrt(3) - y^2));
%Volume of the puddle
V0 = int(b,y,-sqrt(4/3-2/sqrt(3)),sqrt(4/3-2/sqrt(3)));
V = vpa(V0)
```

0.25553393767920640581899176181759 + 0.00000000000000000000000014120312922006717511927170824155i

##

The real part has promising-looking digits, except they're too large by a factor of 10. Any suggestions on what I'm doing wrong? Thanks.

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