[Math] Volume of Viviani’s Window

calculusdefinite integralsgeometryintegrationreal-analysis

I’m taking Real Analysis and I have an assignment to do on calculating the volume of Viviani’s window or dome.

I have to solve this for the sphere $$x^2+y^2+z^2=4$$ and the cylinder $$(x-1)^2+y^2=1$$

Here's a visual representation of the problem:

enter image description here

Here's a figure obtained from the intersection of the 2 figures. (I need to find the volume of that figure.) The image shows a curve which does not have a volume but to understand what volume the problem is asking I imagine as if on the inside it is filled with some material

enter image description here

My problem is that all I have seen on the Internet about this problem involves multivariable calculus, parametrization, polar coordinates, etc. I have not seen any of these yet, so I have to do this problem with integrals in only one variable.

The formulas we have seen in class about the applications of the Riemann Integral are the following:

  1. Arc length L of a curve
    $$L= \int_{a}^{b} \sqrt{1+(f’(x))^2} dx$$

  2. Volume
    $$V_1=\int_{a}^{b} \pi (f(x)^2 dx$$
    $$V_2=\int_{a}^{b} A(x)dx$$
    A(x) is the area of a section of the figure

  3. Lateral area
    $$\int_{a}^{b}2\pi f(x) \sqrt{1+f’(x)}dx$$

I thought on expressing the sphere equation and the cylinder equation in terms of the same variable and integrating. The reason behind this was that Viviani’s dome is composed of the points that belong to the cylinder and sphere at the same time, but this reasoning has failed and I don’t know why.

Any help is very much appreciated and I’m sorry for my broken English.

EDIT

This is all I have for now. Need help for A(x). I don’t know how to treat x as a parameter inside the integral and at the bounds of the integral.

This is all I have for now (need help for A(x))

EDIT 2

@Ertxiem I finally solved the integral and got the final answer:
$$\frac{16}{9}(3\pi-4)$$

I still have a few questions though:

  1. When you said: “You can start by thinking about the base.” what did you really meant by the base? I can’t really see what the base of a figure like that would be, since it’s a curve-like figure, how would you visualize that? If you meant the base of the figure, then I understand that studying it is necessary in order to compute the volume but at what point that computation was used?

  2. What is exactly b(x) and why is it important?

  3. Regarding the height, A(x) is the integral of a function, which is the height taking x as a parameter and y as a variable but, the graph of the function is in the first quadrant, so, doesn’t this mean that we are actually integrating height/2?

Conclusion:

Find the base, find the height, find the area of a section by integrating the previous results, find the volume by integrating the area of a section.

Please point out any flaws, errors. Any suggestions for a better explaining, understanding of the problem are welcome. Thanks!

Best Answer

Here is a detailed expalantion of @DanielWainfleet's idea. Let $\mathcal{V}$ denote the region. Then the the intersection of $\mathcal{V}$ and the plane $x = x_0$ is described by

$$ \mathcal{I}(x_0) \quad : \quad y^2 + z^2 \leq 4 - x_0^2 \quad \text{and} \quad y^2 \leq 1 - (1 - x_0)^2.$$

Intersection

The area of $\mathcal{I}(x_0)$ can be computed by decomposing this into 4 congruent wedges and two congruent isosceles as in the following figure.

Intersection

Indeed, note that the four corners of $\mathcal{I}(x_0)$ are given by $(\pm \sqrt{x_0(2-x_0)}, \pm \sqrt{2(2 - x_0)})$, which follows from solving the system of equations $y^2 + z^2 = 4-x_0^2$ and $y^2 = 1 - (1-x_0)^2$. From this, the angle of each of four wedges is given by $\arctan(\sqrt{x_0/2})$, and so, the area of $\mathcal{I}(x_0)$ is given by

$$ S(x_0) := 2(4 - x_0^2) \arctan(\sqrt{x_0 / 2}) + 2 \sqrt{2x_0}(2 - x_0) $$

Then the volume of $\mathcal{V}$ is obtained by integrating this with respect to $x_0$ over $[0, 2]$. Then,

\begin{align*} \text{[volume of $\mathcal{V}$]} &= \int_{0}^{2} S(x) \, \mathrm{d}x \\ &= \int_{0}^{2} \left[ 2(4 - x^2) \arctan\left(\sqrt{\frac{x}{2}}\right) + 2 \sqrt{2x}(2 - x) \right] \, \mathrm{d}x \\ &= \left[ \frac{2}{3} (4-x) (x+2)^2 \arctan\left(\sqrt{\frac{x}{2}}\right) - \frac{2}{9} \sqrt{2x} \left(3x^2 - 10x + 24\right) \right]_{0}^{2} \\ &= \frac{16 \pi}{3} - \frac{64}{9}. \end{align*}

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