Simplify this problem

Mathematics

1 answer:

ikadub [295]3 years ago

7 0

Answer:

$\boxed{ \frac{ \sqrt[3]{ {x}^{11} } }{4} }$

Step-by-step explanation:

$= > \frac{ {x}^{4} }{ \sqrt[3]{64x} } \\ \\ = > \frac{ {x}^{4} }{ {(64x)}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4} }{ ({64}^{ \frac{1}{3} } )\times ({x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({( {4}^{3} )}^{ \frac{1}{3} }) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({4}^{ \cancel{3} \times \frac{1}{ \cancel{3}} } ) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{4 {x}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4 - \frac{1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{12 - 1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{11}{3} } }{4} \\ \\ = > \frac{ \sqrt[3]{ {x}^{11} } }{4}$

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$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
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and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

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$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$