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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 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  It does not state that anywhere in the problem. Just remember, Mathematics is "casesensitive." Anyway, you have the bottom part of your post marked as "Solution." Do you need help in writing the limits on the integral? The nontrivial 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 nonnegative. 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 xyplane, integrating z over both nappes gives a result, without the necessity of actually integrating, of 0. 

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calculating, integral, triple, volume 
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