2
$\begingroup$

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?)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.