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 Pre-Calculus Pre-Calculus Math Forum

 September 2nd, 2017, 12:12 PM #1 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Volume integration What could be the volume of a solid bounded by $\displaystyle x = y^{2}$ $\displaystyle 4-x = y^{2}$ $\displaystyle z =0\hspace {2mm} and \hspace {2mm} z=3$ September 2nd, 2017, 12:38 PM   #2
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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.

$I=\displaystyle 3\int_{-\sqrt{2}}^{\sqrt{2}}\int_{y^2}^{2}~dx~dy + 3 \int_{-\sqrt{2}}^{\sqrt{2}}\int_{2}^{4-y^2}~dx~dy = 16 \sqrt{2}$
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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: 901 Thanks: 61 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
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Quote:
 Originally Posted by MATHEMATICIAN Thanks, had plotted in a paper. What is that software that you are using? I was getting the wrong anwer as I had not separated the two blocks. By the way, why do we do that?
software is Mathematica. It can be had pretty cheaply by students.

Actually now that I think about it you don't have to separate the two integrals.

$I =3 \displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} \int_{y^2}^{4-y^2}~dx~dy$

works just as well. September 2nd, 2017, 01:07 PM   #5
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Quote:
 Originally Posted by romsek $I =3 \displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} \int_{y^2}^{4-y^2}~dx~dy$ works just as well.
How would you do it if you had to replace variable "y" limits with variable "x" limits? September 2nd, 2017, 01:32 PM   #6
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Quote:
 Originally Posted by MATHEMATICIAN How would you do it if you had to replace variable "y" limits with variable "x" limits?
To do this you would have to break it into 2 integrals, or again just recognize the symmetry.

$I = \displaystyle 3\int_0^2 \int_{-\sqrt{x}}^{\sqrt{x}}~dx ~dy + 3\int_2^4 \int_{-\sqrt{4-x}}^{\sqrt{4-x}}~dx~dy = 6 \int_0^2 \int_{-\sqrt{x}}^{\sqrt{x}}~dx ~dy = 16\sqrt{2}$ September 2nd, 2017, 01:34 PM #7 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 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: 901 Thanks: 61 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
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Quote:
 Originally Posted by MATHEMATICIAN 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?
I would interpret this question as finding the line integral of the field given by

$\textbf{F}=(z, x, y)$

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: 901 Thanks: 61 Math Focus: सामान्य गणित So, the curve is an ellipse? And what about this orientation thing? What is its significance? 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 MATHEMATICIAN Calculus 13 September 1st, 2017 11:12 AM xl5899 Calculus 2 December 10th, 2015 09:09 AM jiasyuen Calculus 9 March 29th, 2015 08:37 PM landonjones6 Calculus 1 February 21st, 2012 02:33 AM izseekzu Calculus 1 January 26th, 2010 06:34 PM

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