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 April 5th, 2018, 02:23 AM #1 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 Calculating Volume using Triple Integral **I am trying to solve this problem but I am having difficulties to finish it. I would appreciate of someone can advice me on how to continue** **Problem:** Calculate $$\iiint_{V} Z\mathrm dV$$ where E is defined by $$x^2+y^2 \le z^2$$and$$x^2+y^2=z^2 \le R^2 with R\gt0$$ **Solution** Using Cylindrical Coordinates $$\iiint_{V} Z\mathrm dV = \iiint_{S} Z\mathrm rdrdzd\theta$$ $$\iiint_{V} Z\mathrm dV = \iiint_{S} Z\mathrm rdrdzd\theta$$ $$x = rcos\theta, y= rsin\theta$$ Last edited by sosoebot; April 5th, 2018 at 02:30 AM. April 5th, 2018, 03:47 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 What is "Z"? Do you mean just the z coordinate? What are "V" and "S"? You don't define them anywhere. April 5th, 2018, 03:51 AM #3 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 Sorry V=E, and it is defined in the problem April 5th, 2018, 11:58 PM #4 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 I have reposted the question here. I made alot of typo errors in the previous post **I am trying to solve this problem but I am having difficulties to finish it. I would appreciate of someone can advice me on how to continue** **Problem:** Calculate $$\iiint_{V} Z\mathrm dV$$ where V is defined by $$x^2+y^2 \le z^2$$and$$x^2+y^2+z^2 \le R^2 with R\gt0$$ **Solution** Using Cylindrical Coordinates $$\iiint_{V} Z\mathrm dV = \iiint_{V} Z\mathrm rdrdzd\theta$$ $$\iiint_{V} Z\mathrm dV = \iiint_{V} Z\mathrm rdrdzd\theta$$ $$x = rcos\theta, y= rsin\theta$$ April 6th, 2018, 04:02 PM #5 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 You still have one unanswered question. Does Z=z? April 6th, 2018, 05:34 PM #6 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 Yes, Z=z as stated in the problem April 6th, 2018, 08:32 PM   #7
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Quote:
 Originally Posted by sosoebot Yes, Z=z as stated in the problem
It does not state that anywhere in the problem.

Just remember, Mathematics is "case-sensitive."

Anyway, you have the bottom part of your post marked as "Solution." Do you need help in writing the limits on the integral? The non-trivial part of the limits is the restriction on the z coordinate. z would have to go from 0 to $\displaystyle \sqrt{x^2 + y^2}= \sqrt{R^2 - r^2}$ due to your inequality. Thus the integral would be
$\displaystyle \int _{0}^{2 \pi} \int _{0}^{R} \int _{0}^{\sqrt{R^2 - r^2}} z ~ r ~ dr ~ dz ~ d \theta$

-Dan

Last edited by topsquark; April 6th, 2018 at 08:40 PM. April 7th, 2018, 05:45 AM #8 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 Thank you very much. I have done the integration and I ended up with $$\frac{\pi R^4}{4}$$ April 7th, 2018, 10:00 AM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I don't see anywhere in the stated problem that z must be non-negative. The region of integration is that portion of the cone, $\displaystyle x^2+ y^2\le z^2$ that lies inside the sphere $\displaystyle x^2+ y^2+ z^2\le R^2$. Taking the square root ignores the lower nappe of the cone. But since the cone is symmetric about the xy-plane, integrating z over both nappes gives a result, without the necessity of actually integrating, of 0. Thanks from topsquark Tags calculating, integral, triple, 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 sosoebot Calculus 2 April 3rd, 2018 12:43 PM zollen Calculus 1 October 8th, 2017 03:34 PM zollen Calculus 8 May 2nd, 2017 02:38 PM zollen Calculus 8 April 9th, 2017 09:51 AM triplekite Calculus 1 December 5th, 2012 08:16 AM

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