Let $$R=\{(x,y,z):y^2+z^2\leq 1\,\, \text{and}\,\, x^2+z^2\leq 1\}.$$

  1. Compute the volume of $R$.
  2. Compute the area of its boundary $\partial R$.

I'm fine with #1. For #2, I have a solution here, which I'm not sure is correct (but I trust the error is with me). I'm wondering where is the error in my method. Consider a cylinder moving left to right along the $y$-axis. Cut it in half along the $yz$-plane, and parameterize it by $y,z$. Let $$D = \{(y,z) \mid -1\leq z\leq 1\}.$$ This infinitely long strip is our domain of parameterization, and the (back half of the) cylinder is parameterized $$C(y,z)= (\sqrt{1-z^2}, y,z).$$

Now to find the surface area across a bounded region $\Delta \subset D$, we would integrate $$\iint_\Delta |n(y,z)|dydz.$$ I get $|n(y,z)|=\frac{1}{\sqrt{1-z^2}}.$ Now to find the surface area (of the back half of intersection between the two cylinders) we can just integrate $$\iint_\Delta \frac{1}{\sqrt{1-z^2}}dydz,$$ where $\Delta$ is the unit disk in the $yz$-plane. I calculate this to be $4$.

But my solution is off by a factor of 2 (according to the answer above, the total integral should be $16$, meaning the integral of the back half should be $8$). What have I failed to consider? (Or perhaps the solution I linked to is off by a factor of 2?)


There are two cylinders $A$ and $B$. The surface of $A\cap B$ consists of two parts:

  1. $(\partial A)\cap B$
  2. $A\cap (\partial B)$

You dealt with only one of the above. So, after doubling your half-cylinder to cylinder, you should double again.


$$ \iint_\Delta \frac{1}{\sqrt{1-z^2}}\mathrm{d}y\mathrm{d}z=2$$ Why? Because $$ \iint_\Delta \frac{1}{\sqrt{1-z^2}}\mathrm{d}y\mathrm{d}z=\int\limits_0^1\int\limits_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \frac{1}{\sqrt{1-z^2}}\mathrm{d}y\mathrm{d}z$$$$=\int\limits_0^1\frac{1}{\sqrt{1-z^2}}y|_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\mathrm{d}z=\int\limits_0^1\frac{1}{\sqrt{1-z^2}} [\sqrt{1-z^2} - (-\sqrt{1-z^2})] \mathrm{d}z$$$$=\int\limits_0^1 2 \mathrm{d}z=2$$


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