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 September 2nd, 2017, 12:12 PM #1 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 Thanks: 60 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: 878 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
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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: 2,122 Thanks: 1102 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: 878 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
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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: 878 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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