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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: 6,786 Thanks: 708 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

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