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September 1st, 2017, 09:57 AM   #11
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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, 09:59 AM   #12
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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, 10:34 AM   #13
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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, 12:12 PM #14 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 911 Thanks: 64 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 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 xl5899 Calculus 2 December 10th, 2015 10:09 AM jiasyuen Calculus 9 March 29th, 2015 09:37 PM PhysicsWizard Calculus 2 March 25th, 2014 07:47 AM pjlloyd100 Calculus 3 February 18th, 2013 12:51 PM izseekzu Calculus 1 January 26th, 2010 07:34 PM

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