My Math Forum Volume integration

 Calculus Calculus Math Forum

September 1st, 2017, 08:57 AM   #11
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,156
Thanks: 731

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by MATHEMATICIAN thanks. i get the same what about the surface integration over the curved surfaces, taking integration function $\displaystyle F\cdot \hat{n}$ where, $\displaystyle \hat{n}$ is unit vector perpendicular to the curved surface i.e. $\displaystyle \frac{x\vec{i} + y\vec{j}}{\sqrt{2}}$
Nope, I'm going to use $\displaystyle n = \hat{r}$ for the curved surfaces and $\displaystyle n = \hat{k}$ or $\displaystyle -\hat{k}$ for the ends.

$\displaystyle F = 4x \hat{i} - 2y^2 \hat{j} + z^2 \hat{k}$

$\displaystyle F = 4r\cos\theta (\cos\theta + \sin\theta) \hat{r} + 2 r^2 \sin^2\theta(\sin\theta - \cos\theta)\hat{\theta} + z^2 \hat{z}$

Curved part (at r=2):
$\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^3_0 4r\cos\theta(\cos\theta + \sin\theta) r d\theta dz$

$\displaystyle = \int^{2\pi}_0 \int^3_0 16\cos\theta(\cos\theta + \sin\theta) d\theta dz$

$\displaystyle = \int^{2\pi}_0 \int^3_0 (16\cos^2\theta + 16 \cos\theta \sin\theta) d\theta dz$

$\displaystyle = \int^{2\pi}_0 \int^3_0 (8\cos 2\theta + 8 + 8\sin 2\theta) d\theta dz$

$\displaystyle = \int^3_0 \left[4\sin 2\theta + 8\theta - 4\cos 2\theta\right]^{2\pi}_0 dz$

$\displaystyle = \int^3_0 (16 \pi) dz$

$\displaystyle = \left[16 \pi z\right]^3_0$

$\displaystyle = 48\pi$

Bottom face (z=0): $\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^2_0 -z^2 r dr d\theta = 0$

Top face (z=3): $\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^2_0 z^2 r dr d\theta = \int^{2\pi}_0 \int^2_0 9r dr d\theta$

$\displaystyle = \int^2_0 \left[9r\theta\right]^{2\pi}_0 dr$

$\displaystyle = \int^2_0 18\pi r dr$

$\displaystyle = \left[9\pi r^2 \right]^2_0$

$\displaystyle = 36\pi$

Therefore, the total of the integral is $\displaystyle 48\pi + 36 \pi = 84 \pi$

September 1st, 2017, 08:59 AM   #12
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,156
Thanks: 731

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by agentredlum [EDIT] Benit13 beat me to it. Took me 2 hours to type this in LaTeX and proof read it [ENDEDIT]
No problem, it's always nice to have a second pair of eyes on the same thing

September 1st, 2017, 09:34 AM   #13
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 901
Thanks: 61

Math Focus: सामान्य गणित
Quote:
 Originally Posted by Benit13 Nope, I'm going to use $\displaystyle n = \hat{r}$ for the curved surfaces and $\displaystyle n = \hat{k}$ or $\displaystyle -\hat{k}$ for the ends. $\displaystyle F = 4x \hat{i} - 2y^2 \hat{j} + z^2 \hat{k}$ $\displaystyle F = 4r\cos\theta (\cos\theta + \sin\theta) \hat{r} + 2 r^2 \sin^2\theta(\sin\theta - \cos\theta)\hat{\theta} + z^2 \hat{z}$ Curved part (at r=2): $\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^3_0 4r\cos\theta(\cos\theta + \sin\theta) r d\theta dz$ $\displaystyle = \int^{2\pi}_0 \int^3_0 16\cos\theta(\cos\theta + \sin\theta) d\theta dz$ $\displaystyle = \int^{2\pi}_0 \int^3_0 (16\cos^2\theta + 16 \cos\theta \sin\theta) d\theta dz$ $\displaystyle = \int^{2\pi}_0 \int^3_0 (8\cos 2\theta + 8 + 8\sin 2\theta) d\theta dz$ $\displaystyle = \int^3_0 \left[4\sin 2\theta + 8\theta - 4\cos 2\theta\right]^{2\pi}_0 dz$ $\displaystyle = \int^3_0 (16 \pi) dz$ $\displaystyle = \left[16 \pi z\right]^3_0$ $\displaystyle = 48\pi$ Bottom face (z=0): $\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^2_0 -z^2 r dr d\theta = 0$ Top face (z=3): $\displaystyle \int \int F \cdot n ~dS = \int^{2\pi}_0 \int^2_0 z^2 r dr d\theta = \int^{2\pi}_0 \int^2_0 9r dr d\theta$ $\displaystyle = \int^2_0 \left[9r\theta\right]^{2\pi}_0 dr$ $\displaystyle = \int^2_0 18\pi r dr$ $\displaystyle = \left[9\pi r^2 \right]^2_0$ $\displaystyle = 36\pi$ Therefore, the total of the integral is $\displaystyle 48\pi + 36 \pi = 84 \pi$
Thanks for reply, I finally started studying after a long time and have lots of queries. I appreciate your help.

Now m wondering how to find normal to a surface
$\displaystyle z =1-x^{2}$
$\displaystyle 0\leq x \leq 1$
$\displaystyle -2 \leq y \leq 2$

 September 1st, 2017, 11:12 AM #14 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित I got $\displaystyle 2t\vec {i} + \vec{k}$ where, $\displaystyle 0\leq t \leq 1$ as perpendicular to the surface, is it correct?

 Tags integration, volume

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post xl5899 Calculus 2 December 10th, 2015 09:09 AM jiasyuen Calculus 9 March 29th, 2015 08:37 PM PhysicsWizard Calculus 2 March 25th, 2014 06:47 AM pjlloyd100 Calculus 3 February 18th, 2013 11:51 AM izseekzu Calculus 1 January 26th, 2010 06:34 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top