What is the volume of intersection of the three cylinders with axes of length $1$ in $x, y, z$ directions starting from the origin, and with radius $1$?
How to compute the volume of this object via integration?
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0The title could be more descriptive – 2010-07-30
3 Answers
The answer can be found on pages 27-28 of the following link.
http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf
Pictured is one sixteenth of the overall shape.
Note that the top face of the shape is a piece of $x^2+z^2=1$ and the front face is a piece of $x^2+y^2=1$; I’ve drawn in the edge where these two faces meet.
To find its volume, we’ll integrate cross-sectional area with respect to $x$ (trial and error shows that this is vastly simpler than doing it with respect to $y$.)
We’ll use $A(x)$ to denote the area of the cross-section at $x$. This is a piecewise-defined function. The rectangle in the drawing is the location of $A$’s discontinuity. For $x \leq \sqrt{2}/2$, the cross-section is a rectangle with width $x$ and height $\sqrt{1-x^2}$, thus $A(x)=x\sqrt{1-x^2}$. For $x \geq \sqrt{2}/2$, the cross-section is a square of side-length $\sqrt{1-x^2}$, giving $A(x)=1-x^2$.
Thus we get $$V = 16\left(\int_0^{\sqrt{2}/2}x\sqrt{1-x^2}\,dx+\int_{\sqrt{2}/2}^1(1-x^2)\,dx\right)=16-8\sqrt{2}.$$
Use triple integration. $V=\int \int \int dx dy dz$
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0It is clear that this triple integral would give the volume, but I think you should add some details about how to actually compute it. – 2016-08-15