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Natasha2012 [34]
2 years ago
10

Express the integral as an iterated integral in six different ways, where E is the solid bounded by y=4-x^2-4z^2 and y=0

Mathematics
1 answer:
zmey [24]2 years ago
5 0
Assuming you need the integral expressing the volume of E, the easiest setup is to integrate with respect to y first.

This is done with either

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dx\,\mathrm dz
\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dz\,\mathrm dx

Thanks to symmetry, integrating with respect to either x or z first will be nearly identical.

First, with respect to x:

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz
\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dz\,\mathrm dy

Next, with respec to z:

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dy\,\mathrm dx
\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dx\,\mathrm dy
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