The largest area: Part II

Here we will consider the 3D analogue of the problem mentioned in the previous post, and first define the two paraboloids (the 3D analogue of parabolas):

$\begin{aligned} z_1\left(x,y\right) &= x^2 + y^2 + 1, \quad z_2\left(x,y\right) &= x^2 + y^2 \end{aligned}$

This time we’ll call the tangent point $T = \left(x_T,y_T\right)$ . Then the slope of the tangent plane of $z_1\left(x,y\right)$ at $T$ in each direction is given by the partial derivatives of $z_1$ :

$\displaystyle \frac{\partial z_1}{\partial x}\left(x_T,y_T\right) = 2x_T, \qquad \displaystyle \frac{\partial z_1}{\partial y}\left(x_T,y_T\right) = 2y_T$

And so the equation for the tangent plane $P$ is constructed by taking $z_1(x_T,y_T)$ and adding linear functions with these slopes for the two coordinates $x$ and $y$ :

$P\left(x,y;T\right) = 2x_T\left(x - x_T\right) + 2y_T\left(y - y_T\right) + x_T^2 + y_T^2 + 1$

Play around with the sliders here to see the tangent plane and the associated volume underneath change shape (Note that I had to strategically cut the outer paraboloid so we can see what’s going on).

If you solve for the region of intersection between $P\left(x,y;T\right)$ and $z_2\left(x,y\right)$ , and find that they intersect at the circle of radius $1$ centred at $T$ . Let’s call this circle $R$ .

From here, we can define the volume $V$ bounded below $P\left(x,y;T\right)$ and above $z_2\left(x,y\right)$ as the double integral:

$V = \displaystyle \iint_R P\left(x,y;T\right) - z_2\left(x,y\right) \mathrm{d}R = \displaystyle \iint_R 1 - \left(x - x_T\right)^2 - \left(y - y_T\right)^2 \mathrm{d}R$

We now introduce a change of variables $u = x + x_T$ and $v = y + y_T$ . This transformation is linear and the Jacobian is 1, so our region of intersection simply becomes a circle of radius $1$ centered at the origin. Thus $V$ becomes

$V = \displaystyle \iint_{R'} \displaystyle \left(1 - u^2 - v^2\right) \mathrm{d}R'$

We first note that $V$ is (once again, if you read the first post in this series) indifferent to the tangent point $T$ ! To compute this integral, we now transform our coordinates from Cartesian to Polar coordinates:

$u = r \cos\left(\theta\right), \quad v = r \sin\left(\theta\right) \longrightarrow \mathrm{d}R' = r \, \mathrm{d}r \, \mathrm{d}\theta,$

a common trick from multivariable calculus. Then we have

$\begin{aligned} V &= \iint_{R'} \displaystyle \left(1 - u^2 - v^2\right) \mathrm{d}R' \\ &= \displaystyle \int_{-\pi}^{\pi} \int_0^1 \left(1 - r^2\right) r \, \mathrm{d}r \, \mathrm{d}\theta \\ &= \displaystyle \int_{-\pi}^{\pi} 1 \, \mathrm{d}\theta \int_0^1 \left(1 - r^2\right) r \, \mathrm{d}r \\ &= \left[\theta\right]_{-\pi}^{\pi} \left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 \\ &= 2\pi \cdot \frac{1}{4} \\ V& = \frac{\pi}{2} \end{aligned}$

So just like the base case with the parabolas, in this case the volume is a constant. Is this true in all higher dimensions, for all possible $n$ -dimensional paraboloids? This will be discussed in the next post, but [spoiler], yes. The integrals always turn out to be independent of where you place the tangent point.

This in itself is pretty cool, but something that really stood out to us was the pattern of values themselves. If we call the area, volume or hyper-volume under the tangent in $n$ dimensions $S_n$ , and we call the base case $n=1$ because parabolas are 1D curves, then we have

$S_1 = \displaystyle \frac{4}{3}, \quad S_2 = \displaystyle \frac{\pi}{2} , \quad S_3 = \text{?}$

Can you guess the next value in the series?

Even though we only have two terms, they strongly reminded me of something I read about as an undergrad, and David was able to provide the calculations to prove the conjecture!

See the third and final post in this series to see the conclusion to the story, it’ll be a ball, I promise.

Stay curious!

2 responses to “The largest area: Part II”

A problem of parabolas – The Calculus of Explanations

August 15, 2024 at 5:31 pm

[…] What we found was extremely surprising, I hope you’ll join us to explore this problem further in the next post. […]

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A problem of hyper-paraboloids – The Calculus of Explanations

August 15, 2024 at 5:37 pm

[…] Part II of this series, we wondered whether the result of the area being constant continued into higher […]

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The Calculus of Explanations