Lake Volume

Calculus Level 3

The height profile of a valley basin can be described by the two-dimensional parabolic function $h(x, y) = \frac{x^2}{144\,\text{m}} + \frac{y^2}{324\,\text{m}}.$ Now the basin is filled by a rainstorm to a height of $h_0 = 12\,\text{m}.$ What is the volume of the resulting lake $($ in $\text{m}^3)$ to the nearest integer?

Hint: Find the shape of the cross-sectional areas enclosed by the equipotential lines $h(x, y) = z = \text{constant}$ . The volume then results from the integral $\displaystyle V = \int_0^{h_0} A (z)\, dz$ over the cross-sectional area $A(z).$

The answer is 48858.

2 solutions

Chew-Seong Cheong
Nov 29, 2017

For a standard ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ , the area of the ellipse is $A(1) = \pi ab$ . Therefore, $A(z) = \pi abz$ and the volume

$\begin{aligned} V & = \int_0^{h_0} A(z) \ dz = \int_0^{h_0} \pi abz \ dz = \frac {\pi a b z^2}2 \bigg|_0^{h_0} = \frac {\pi \cdot \sqrt{144} \cdot \sqrt{324}\cdot 12^2}2 \approx \boxed{48858} \end{aligned}$

Moderator note:

A simple change in perspective can save quite a bit of work!

I dissent if h is measure in meter like a and b, how come an area have a dimension of a volume, there is something that not fits

Mara Jares - 3 years, 6 months ago

The $z$ in the formula $A(z) = \pi abz$ is a number. It is $\dfrac {z \text{ m}}{1 \text{ m}}$ . Note that $A(1) = \pi ab \text{ m}^2$ . 1 is a number here, therefore, $z$ is a number.

Chew-Seong Cheong - 3 years, 6 months ago

But the upper limit of the integral is ho what as per the enunciate is 12 meter. I think the area is pi a(z) b(z), sosemiaxis of the ellipse are the function of z. Multiplying by z you are actually obtaining the first moment to the bottom of the basin.

Mara Jares - 3 years, 6 months ago

@Mara Jares – I don't quite understand what you are saying. But the approach by Markus and I came to the same answer which has been confirmed by the Challenge Master. I hope you can find out the reason yourself.

Chew-Seong Cheong - 3 years, 6 months ago

@Chew-Seong Cheong – Sorry but I do not understand why you asume values for a as 12 and for b=18 for z=1. By doing so At level z=12 according to the equation our ellipse will have semiaxis smaller what it is inconsistent with the reality.

Mara Jares - 3 years, 6 months ago

@Mara Jares – Sorry, I didn't explain clearly earlier. I also didn't check the question again earlier. After reading the question again. I recall I solved the problem this way because of the hint. If you read the question properly the hint mentions that "the equipotential lines $h(x, y) = z = \text{constant}$ . It clearly told me that $z$ is a constant which is a dimensionless quantity just as I mentioned in my first explanation to you and it is exactly opposite to what you have assumed that it has a dimension of length which is meter in SI unit.

Since $z$ in $z = \dfrac {x^2}{\color{#3D99F6}a^2} + \dfrac {y^2}{\color{#D61F06}b^2} = \dfrac {x^2}{\color{#3D99F6}144}+ \dfrac {y^2}{\color{#D61F06}324}$ is a constant, it means that the dimensions $x$ and $y$ of the ellipse which is given by the RHS $\dfrac {x^2}{\color{#3D99F6}a^2} + \dfrac {y^2}{\color{#D61F06}b^2}$ is directly proportional to the LHS $z$ . That is when $z=1$ (no unit), the area of a standard ellipse $\dfrac {x^2}{\color{#3D99F6}a^2} + \dfrac {y^2}{\color{#D61F06}b^2}=1$ is $A(1) = \pi \color{#3D99F6}a \color{#D61F06}b$ , then area of ellipse for any $z$ is $A(z) = \pi {\color{#3D99F6}a}{\color{#D61F06}b}z$ .

By now, I hope you can see that $\color{#3D99F6}a = 12$ and $\color{#D61F06}b = 18$ are not assumption but that $\color{#3D99F6} a^2 = 144 \implies a = \sqrt{144} = 12$ and $\color{#D61F06} b^2 = 324 \implies b = \sqrt{324} = 18$ .

I hope that you can understand now.

Chew-Seong Cheong - 3 years, 6 months ago

There is no mistake: $a$ and $b$ are not measured in meters, their squares are, which means thet the product $ab$ is meters and not meters squared.

Steve Gualtieri - 3 years, 4 months ago

But if dimensions of a^2 is meter and also b^2 is also meter, means that area of ellipse Pi times ab is a fourth power isn't it.

Mara Jares - 3 years, 3 months ago

@Mara Jares – No: let's use $[x]$ to denote the dimension of a quantity. $1$ means that the quantity is a constant, as in $[\pi]=1$ .

$[a^2]=m\Rightarrow[a]=m^{\frac{1}{2}}$

Same goes for $b$ , while $[z]=m$ therefore:

$[A(z)]=[\pi abz]=[\pi][a][b][z]=1\cdot m^{\frac{1}{2}}\cdot m^{\frac{1}{2}}\cdot m=m^2$

Steve Gualtieri - 3 years, 3 months ago

@Steve Gualtieri – But Steve as far as I know, the semi axis of an ellipse is a length, feet, inches meter or millions of kilometers like distant from earth to sun at perihelia for instance. Not square root of length. The volume V is the integral of dV= A(z)dz being A(z) the ellipse area at section z. Therefore A(z) = Pie * a(z) b(z). But not Pie times a(z) b(z)*z

Mara Jares - 3 years, 3 months ago

@Mara Jares – You are $100\%$ right (keep in mind though that the perihelion is not the lenght $a$ of the major semiaxis of earth's orbit but rather the distance $a-c$ , where $c=\sqrt{a^2-b^2}$ is the distance between the center of the ellipsis and one of it's focuses), and I had to give a thought as to why both what you said and Chew-Song Cheong's proof are valid. In the end, here is some more explanation as to why those formula work and there is no error in the formulation of the problem.

In the equation $h(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ , the profile of the lake at height $z$ is given by the equation $\frac{x^2}{a^2z}+\frac{y^2}{b^2z}=1$ . This implies that it is an ellpisis with semiaxes $a\sqrt{z}$ and $b\sqrt{z}$ , and the area of this shape is of course $A(z)=\pi abz$ .

Now, since $a^2$ and $b^2$ are measured in meters, $z$ is measured in meters as well. This is necessary and coherent as $z=h(x,y)$ is defined as a height, and with these dimensions we have $[A(z)]=m^2\,\forall z$ . When you plug in $z=1$ you get $A(1)=\pi ab$ , which looks like it lacks a "meter" and is therefore a lenght, but the thruth is that the missing meter is hidden with the disappearing $z$ , which is still measured in meters.

Of course, if we measured $a$ and $b$ in meters, as you say we should, all of the formula would still be valid since $z$ would be adimensional, but the problem would not make sense because an height cannot be adimensional.

tl;dr: the semiaxis of an ellipse are measured in meters, but $a$ and $b$ are only semiaxes of the ellipse at height 1, and the dimension is still coherent because they correspond to $a\sqrt{z}$ and $b\sqrt{z}$ for $z=1$ , and $z$ is measured in meters as well.

Steve Gualtieri - 3 years, 3 months ago

Markus Michelmann
Nov 6, 2017

The height profile is given by $h(x,y) = \alpha x^2 + \beta y^2$ with constants $\alpha = 1/12^2$ and $\beta = 1/18^2$ . The lines $h(x,y) = z = \text{const}$ enclose ellipses: $\begin{aligned} \frac{h(x,y)}{z} &= \frac{\alpha x^2}{z} + \frac{\beta y^2}{z} = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\ \Rightarrow \quad y &= b\sqrt{1 - \frac{x^2}{a^2}} \end{aligned}$ with the half-axes $a = \sqrt{z/\alpha} = 12 \sqrt{z}$ , $b = \sqrt{z/\beta} = 18 \sqrt{z}$ . The area of ellipse results to $\begin{aligned} A(z) &= \int_{-a}^{a} b\sqrt{1 - \frac{x^2}{a^2}} dx & &| x = a \sin \phi \\ &= ab \int_{-\pi}^{\pi} \sqrt{1 - \sin^2 \phi} \cos \phi \,d\phi = ab \int_{-\pi}^{\pi} \cos^2 \phi\, d\phi \\ &= \frac{1}{2} ab \int_{-\pi}^{\pi} (\sin^2 \phi + \cos^2 \phi) d\phi = \frac{1}{2} ab \int_{-\pi}^{\pi} d\phi \\ &= \pi a b = \frac{\pi z}{\sqrt{\alpha \beta}} \end{aligned}$ Therefore, the lake volume is calculated by the integral $V = \int_{0}^{h_0} A(z) dz = \frac{\pi}{\sqrt{\alpha \beta}} \int_0^{h_0} z dz = \frac{\pi h_0^2}{2 \sqrt{\alpha \beta}} = \frac{1}{2} \pi a b h_0 = \frac{1}{2} \pi \cdot 12 \cdot 18 \cdot 12^2 \approx 48858$

You should specify exact value of Pi for calculation. First I used mistakenly relying on my memory 3.1417 value (instead of 3.1416) and got 48860 cubic M as answer with Brilliant reply as 'Incorrect', so I used 3.14 value that returned 48833 cubic M with Brilliant reply again as Incorrect!

Mirek Baudys - 3 years, 6 months ago

I belive your answer is dimensionally wrong since is proportional to the fourth power what cannot be.

Mara Jares - 3 years, 6 months ago

12 and 18 have the units sqrt(m), therefore it should be fine. Look at the equation for the height, it's x^2/144m +..., not x^2/144m^2.

Malte Diederich - 3 years, 6 months ago

Lake Volume

The answer is 48858.

2 solutions

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