September 2nd, 2017, 12:12 PM  #1 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  Volume integration
What could be the volume of a solid bounded by $\displaystyle x = y^{2}$ $\displaystyle 4x = y^{2}$ $\displaystyle z =0\hspace {2mm} and \hspace {2mm} z=3 $ 
September 2nd, 2017, 12:38 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716 
Have you plotted this? $I=\displaystyle 2 \times 3\int_{\sqrt{2}}^{\sqrt{2}}\int_{y^2}^{2}~dx~dy = 16 \sqrt{2}$ The factor of 2 is due to the symmetry of the shape. The factor of 3 is due to the simple integration over $z$ The limits on $y$ are due to where the two surfaces intersect. If you don't like the symmetry you can do it with two integrals. $I=\displaystyle 3\int_{\sqrt{2}}^{\sqrt{2}}\int_{y^2}^{2}~dx~dy + 3 \int_{\sqrt{2}}^{\sqrt{2}}\int_{2}^{4y^2}~dx~dy = 16 \sqrt{2} $ Last edited by romsek; September 2nd, 2017 at 12:41 PM. 
September 2nd, 2017, 12:49 PM  #3 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
Thanks, had plotted in a paper. What is that software that you are using? I was getting the wrong answer as I had not separated the two blocks. By the way, why do we do that? Is it because of two differeny functions?(different relation between x and y in those two portion) Last edited by MATHEMATICIAN; September 2nd, 2017 at 12:59 PM. 
September 2nd, 2017, 12:52 PM  #4  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716  Quote:
Actually now that I think about it you don't have to separate the two integrals. $I =3 \displaystyle \int_{\sqrt{2}}^{\sqrt{2}} \int_{y^2}^{4y^2}~dx~dy$ works just as well.  
September 2nd, 2017, 01:07 PM  #5 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  
September 2nd, 2017, 01:32 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716  Quote:
$I = \displaystyle 3\int_0^2 \int_{\sqrt{x}}^{\sqrt{x}}~dx ~dy + 3\int_2^4 \int_{\sqrt{4x}}^{\sqrt{4x}}~dx~dy = 6 \int_0^2 \int_{\sqrt{x}}^{\sqrt{x}}~dx ~dy = 16\sqrt{2}$  
September 2nd, 2017, 01:34 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716 
There is actually 4 fold symmetry to this shape so the integration limits can be reduced further. I'll leave that to you to play with. hint: $I=12\int_0^2 \int_0^\sqrt{x}~dy~dx$ 
September 2nd, 2017, 01:49 PM  #8 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
One more question: Evaluate: $\displaystyle \int_{C} (z~dx +x~dy+y~dz)$ where, C is the trace of cylinder $\displaystyle x^{2}+y^{2}=1$ in the XY plane, $\displaystyle y+z=2$. Orient C counter clockwise. Is something wrong with this question? 
September 2nd, 2017, 02:05 PM  #9  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716  Quote:
$\textbf{F}=(z, x, y)$ over the curve in the plane $y+z=3$ whose projection onto the $xy$ plane is given by $x^2 + y^2 = 1$  
September 2nd, 2017, 02:09 PM  #10 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
So, the curve is an ellipse? And what about this orientation thing? What is its significance?


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