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September 2nd, 2017, 01:12 PM   #1
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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 $
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September 2nd, 2017, 01:38 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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September 2nd, 2017, 01:49 PM   #3
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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)

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September 2nd, 2017, 01:52 PM   #4
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Quote:
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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.
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September 2nd, 2017, 02:07 PM   #5
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Quote:
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$I =3 \displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} \int_{y^2}^{4-y^2}~dx~dy$

works just as well.
How would you do it if you had to replace variable "y" limits with variable "x" limits?
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September 2nd, 2017, 02:32 PM   #6
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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}$
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September 2nd, 2017, 02:34 PM   #7
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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:

$I=12\int_0^2 \int_0^\sqrt{x}~dy~dx$
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September 2nd, 2017, 02:49 PM   #8
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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?
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September 2nd, 2017, 03:05 PM   #9
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Quote:
Originally Posted by MATHEMATICIAN View Post
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 $x^2 + y^2 = 1$
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September 2nd, 2017, 03:09 PM   #10
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So, the curve is an ellipse? And what about this orientation thing? What is its significance?
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