New Notes Blog: How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and...

Saturday, 6 June 2015

How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and...

First octant means $\displaystyle{x}\geq{0},{y}\geq{0},{z}\geq{0}.$ The formula $\displaystyle{y}=\frac{{27}}{{8}}{{x}}^{{3}}$ may be written as $\displaystyle{y}={{\left(\frac{{3}}{{2}}{x}\right)}}^{{3}}.$ We need the (simple) inverse formulas, too: $\displaystyle{x}=\frac{{2}}{{3}}{\sqrt[{{3}}]{{{y}}}},{z}={1}-\frac{{y}}{{2}}.$

Because $\displaystyle{z}\geq{0}$ we see $\displaystyle{y}\leq{2}$ and therefore $\displaystyle{x}\leq\frac{{2}}{{3}}{\sqrt[{{3}}]{{{2}}}}.$ Also, $\displaystyle{z}={1}-\frac{{y}}{{2}}\leq{1}.$ Now we are ready to write the integral in different orders of variables.

If $\displaystyle{x}$ is fixed, $\displaystyle{y}$ can be between $\displaystyle{{\left(\frac{{3}}{{2}}{x}\right)}}^{{3}}$ and $\displaystyle{2}$ in $\displaystyle{x}{y}$ plane $\displaystyle{z}={0}.$ For fixed $\displaystyle{x}$ and $\displaystyle{y}$ we see $\displaystyle{z}$ can...

$\displaystyle{\int_{{{x}={0}}}^{{{x}=\frac{{2}}{{3}}{\sqrt[{{3}}]{{{2}}}}}}}{\left.{d}{x}\right.}{\int_{{{y}={{\left(\frac{{3}}{{2}}{x}\right)}}^{{3}}}}^{{{y}={2}}}}{\left.{d}{y}\right.}{\int_{{{z}={0}}}^{{{z}={1}-\frac{{y}}{{2}}}}}{\left.{d}{z}\right.}{\left({1}\right)}={\int_{{{x}={0}}}^{{{x}=\frac{{2}}{{3}}{\sqrt[{{3}}]{{{2}}}}}}}{\int_{{{y}={{\left(\frac{{3}}{{2}}{x}\right)}}^{{3}}}}^{{{y}={2}}}}{\int_{{{z}={0}}}^{{{z}={1}-\frac{{y}}{{2}}}}}{1}{\left.{d}{z}\right.}.$

If we start from $\displaystyle{y}$ it can be from $\displaystyle{0}$ to $\displaystyle{2},$ then $\displaystyle{x}$ can be from $\displaystyle{0}$ to $\displaystyle\frac{{2}}{{3}}{\sqrt[{{3}}]{{{y}}}}$ and $\displaystyle{z}$ can be from $\displaystyle{0}$ to $\displaystyle{1}-\frac{{y}}{{2}}.$ The integral is

$\displaystyle{\int_{{{y}={0}}}^{{{y}={2}}}}{\int_{{{x}={0}}}^{{{x}=\frac{{2}}{{3}}{\sqrt[{{3}}]{{{y}}}}}}}{\int_{{{z}={0}}}^{{{z}={1}-\frac{{y}}{{2}}}}}{1}{\left.{d}{y}\right.}{\left.{d}{x}\right.}{\left.{d}{z}\right.}.$

And if we start with $\displaystyle{z},$ it can be from $\displaystyle{0}$ to $\displaystyle{1},$ then $\displaystyle{y}$ can be from $\displaystyle{0}$ to $\displaystyle{2}-{2}{z}$ and $\displaystyle{x}$ from $\displaystyle{0}$ to $\displaystyle\frac{{2}}{{3}}{\sqrt[{{3}}]{{{y}}}},$ i.e.

$\displaystyle{\int_{{{z}={0}}}^{{{z}={1}}}}{\int_{{{y}={0}}}^{{{2}-{2}{z}}}}{\int_{{{x}={0}}}^{{{x}=\frac{{2}}{{3}}{\sqrt[{{3}}]{{{y}}}}}}}{1}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}.$

It is not that simple to sketch $\displaystyle{V}$ but I tried. :)

New Notes Blog

Saturday, 6 June 2015

How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 6 June 2015

How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and...

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?