Volume (Cavalieri)

Question

Determine the volume of $$ M:=\left\{(x,y,z)\in\mathbb{R}^3: z\in [0,2\pi],(x-\cos(z))^2+(y-\sin(z))^2\leq\frac{1}{4}\right\} $$

My idea is to use the principle of Cavalieri, i.e. to set $$ M_z:=\left\{(x,y)\in\mathbb{R}^2: (x,y,z)\in M\right\} $$ and then calculate $$ \operatorname{vol}_3(M)=\int\limits_0^{2\pi}\operatorname{vol}_2(M_z)\, dz $$

So I have to calculate $\operatorname{vol}_2(M_z)$.

I set $a:=x-\cos(z)$ and $b:=y-\sin(z)$ and then the condition $a^2+b^2\leq\frac{1}{4}$ means that $$ \lvert a\rvert\leq\frac{1}{2}\Leftrightarrow \cos(z)-\frac{1}{2}\leq x\leq\cos(z)+\frac{1}{2},~~~~~\lvert b\rvert\leq\frac{1}{4}\Leftrightarrow\sin(z)-\frac{1}{2}\leq y\leq\sin(z)+\frac{1}{2}. $$

So I used Fubini and calculated $$ \operatorname{vol}_2(M_z)=\int\limits_{\sin(z)-\frac{1}{2}}^{\sin(z)+\frac{1}{2}}\int\limits_{\cos(z)-\frac{1}{2}}^{\cos(z)+\frac{1}{2}}1\, dx\, dy=1. $$

But then I get $$ \operatorname{vol}_3(M)=2\pi $$

and the result should be $\frac{\pi^2}{2}$. So where is my mistake?

Answer 1

You are making a significant mistake at the point "So I used Fubini...". Your double integral is over a square, when it should be over a circle. The inner integral should have limits that depend on the outer integral's variable.

Answer 2

You are making a mistake in concluding the following statement: $$ (x-\cos(z))^2+(y-\sin(z))^2\leq\frac{1}{4} \iff \\ \cos(z)-\frac{1}{2}\leq x\leq\cos(z)+\frac{1}{2} \,\,\,\,\, \text{and} \,\,\,\,\, \sin(z)-\frac{1}{2}\leq y\leq\sin(z)+\frac{1}{2} $$ And so you conclude the set $M_z$ is the rectangle $R_z = \left[\cos(z)-\frac{1}{2}, \cos(z)+\frac{1}{2}\right] \times \left[\sin(z)-\frac{1}{2}, \sin(z)+\frac{1}{2}\right]$ and apply Fubini's theorem.

The mistake is $\iff$ in the above statement when in fact it is only true in one direction (the left implies the right but not the other way). To see that the right does not imply the left take $x = \cos(z)+\frac{1}{2}$ and $y = \sin(z)+\frac{1}{2}$. Then clearly $(x, y) \in R_z$, but $(x-\cos(z))^2+(y-\sin(z))^2 = (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{2} > \frac{1}{4}$ which proves $(x, y) \notin M_z$.

So, we can conclude that $M_z \subset R_z$ but not $M_z = R_z$ (in fact $M_z$ is a circle contained in the rectangle $R_z$). So you were integrating over a larger area and you got a larger answer too.