User Name Remember Me? Password

 Calculus Calculus Math Forum

 October 18th, 2017, 01:01 PM #1 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Surface area via integration confusion Hello forum, I am struggling on this problem. I am asked to evaluate $\oint \vec r \dot \,d\vec \sigma$ over the whole surface of the cylinder bounded by $x^2+y^2=1, z=0, z=3$ It seems pretty straight forward geometrically as it is just a unit circle at $z=0$ and then it extends up to $z=3$ forming a cylinder. Initially, I parametrized the unit circle as $\langle cos(t), sin(t)\rangle$ and $d\vec \sigma$ pointing normal in the positive z direction. This did not go anywhere however. Now I am thinking I should just work in cylindrical coordinates, but I am just getting kind of confused working with this. Does anyone have any tips for me? Thank you always! October 18th, 2017, 01:17 PM #2 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics I should add... I did $\iint_A \,dA \Rightarrow \int_{0}^{3} dz \int_{0}^{2\pi} d\theta + 2\pi$ = $6\pi + 2\pi = 8\pi$ However, I did this using different techniques from what was asked in the problem. The $2\pi$ added at the end is to account for the "top circle" of the cylinder. October 18th, 2017, 03:07 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Because the figure is not "smooth", this has to be done in three parts: 1) The base at z= 0. This is a disk with center at (x, y)= (0, 0). We can take "$\displaystyle d\vec{\sigma}$" to be $\displaystyle <0, 0, -1> dxdy$ or, in polar coordinates, $\displaystyle <0, 0, -1>r drd\theta$ ("-1" because it is pointing downward. Integrating $\displaystyle \vec{r}$, which, I assume, is $\displaystyle$(?), That gives $\displaystyle \int_0^{2\pi}\int_0^1 r^2 drd\theta$. 2) The base at z= 3. Now $\displaystyle d\vec{\sigma}= \,<0, 0, 1>dxdy$ or $\displaystyle d\vec{\sigma}= \,<0, 0, 1>r dr d\theta$. The integral is the same as above. The difference between the "-1" and "1" is irrelevant because the integrand is independent of z. 3) The curved sides. Yes, you can take $\displaystyle x= cos(\theta)$, $\displaystyle y= \sin(\theta)$, z= z. $\displaystyle \vec{d\sigma}= \,<\cos(\theta), -\sin(\theta), 0>d\theta dz>$. The integrand, $\displaystyle \vec{r}= \,<\cos(\theta), -\sin(\theta), 0>$ so the integral becomes $\displaystyle \int_0^3\int_0^{2\pi} \cos^2(\theta)+ \sin^2(\theta) d\theta dz= \int_0^3\int_0^{2\pi} d\theta dz$ Thanks from SenatorArmstrong Last edited by skipjack; October 18th, 2017 at 11:25 PM. October 18th, 2017, 04:06 PM   #4
Senior Member

Joined: Nov 2015
From: United States of America

Posts: 198
Thanks: 25

Math Focus: Calculus and Physics
Quote:
 Originally Posted by Country Boy Because the figure is not "smooth", this has to be done in three parts: 1) The base at z= 0. This is a disk with center at (x, y)= (0, 0). We can take "$\displaystyle d\vec{\sigma}$" to be $\displaystyle <0, 0, -1> dxdy$ or, in polar coordinates, $\displaystyle <0, 0, -1>r drd\theta$ ("-1" because it is pointing downward. Integrating $\displaystyle \vec{r}$, which, I assume, is $\displaystyle$(?), That gives $\displaystyle \int_0^{2\pi}\int_0^1 r^2 drd\theta$. 2) The base at z= 3. Now $\displaystyle d\vec{\sigma}= \,<0, 0, 1>dxdy$ or $\displaystyle d\vec{\sigma}= \,<0, 0, 1>r dr d\theta$. The integral is the same as above. The difference between the "-1" and "1" is irrelevant because the integrand is independent of z. 3) The curved sides. Yes, you can take $\displaystyle x= cos(\theta)$, $\displaystyle y= \sin(\theta)$, z= z. $\displaystyle \vec{d\sigma}= \,<\cos(\theta), -\sin(\theta), 0>d\theta dz>$. The integrand, $\displaystyle \vec{r}= \,<\cos(\theta), -\sin(\theta), 0>$ so the integral becomes $\displaystyle \int_0^3\int_0^{2\pi} \cos^2(\theta)+ \sin^2(\theta) d\theta dz= \int_0^3\int_0^{2\pi} d\theta dz$
Thanks for clearing up my confusion!

Last edited by skipjack; October 18th, 2017 at 11:27 PM. October 18th, 2017, 04:24 PM #5 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey stuff. I'm going to add an extra comment here. It has no direct bearing on how to solve the current problem, which has been nicely done. However... When putting this problem into cylindrical coordinates, you really should make the substitution $\displaystyle x = r ~ \cos( \theta )$ and $\displaystyle y = r ~ \sin( \theta )$ which has a differential area element $\displaystyle r~dr~d \theta$. In this case, we are looking at the cylinder bounded by $\displaystyle x^2 + y^2 = 1$ so r = 1 anyway. But I think it is a valuable comment, as this will not always be the case. -Dan Thanks from Country Boy and SenatorArmstrong Last edited by skipjack; October 18th, 2017 at 11:20 PM. Tags area, confusion, integration, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fivestar Calculus 2 September 12th, 2016 06:59 PM mathsandmusic Pre-Calculus 8 March 25th, 2016 12:06 PM Icarus Calculus 4 August 18th, 2012 12:17 PM jackinda Calculus 2 January 12th, 2011 02:22 PM xdeathcorex Calculus 3 October 5th, 2010 07:26 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      