Multiplying and Dividing Rational Numbers

Mathematics

1 answer:

Dafna11 [192]3 years ago

4 0

Answer:

the answer is 9.9

Step-by-step explanation:

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Hey there! :D

$3( g^{3})$

Put 2 in as g.

$3( 2^{3})$

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I hope this helps!
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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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