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Evaluate the integral $$\iiint_\Omega xyz\,\mathrm dx\mathrm dy\mathrm dz$$

where $\Omega$ lies in the first octant $x,y,z\geq0$ between the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+z^2=4$ and above the cone $z^2=\frac{1}{3}(x^2+y^2),z\geq0.$

Give the value of the integral rounding to $2$ decimal places.

Really not sure where to start this question, possibly changing to spherical coordinates?

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We can solve this integral using spherical coordinates. There we have:

$$\begin{cases}x=R\sin\theta \cos\phi\\ y=R\sin\theta \sin\phi\\ z=R\cos\theta\end{cases}$$

Then we want to calculate $$I=\iiint_{\Omega} {R^3\sin^2{\theta}}\sin\phi \cos\phi \cos\theta R^2\sin\theta \,\mathrm d\theta \mathrm d\phi \mathrm dR$$ which can be simplified to:

$$I={1\over 2}\iiint_{\Omega} {R^5\sin^3{\theta}}\sin2\phi \cos\theta \, \mathrm d\theta \mathrm d\phi\mathrm dR$$

where $\Omega$ denotes the volume between $R=1$ , $R=2$ and $0\le\theta\le{\pi\over 3}$ while $0\le\phi\le{\pi\over 2}$. Here after we can convert that integral to below:

$$I={1\over 2}\int_{R=1}^{R=2}\int_{\phi=0}^{\phi={\pi/ 2}}\int_{\theta=0}^{\theta={\pi/ 3}} {R^5\sin^3{\theta}}\sin2\phi \cos\theta\, \mathrm d\theta \mathrm d\phi\mathrm dR$$

and we finally obtain:

$$I={189\over 256}$$

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  • $\begingroup$ Hi, thanks for your response. Just want to clarify some things. I am quite new to spherical coordinates. I am assuming $R^2sin\theta$ comes from the Jacobian with respect to $R,\theta$ and $\phi$? I'm assuming the $\frac{1}{2}$ is taken out due to the $sin\phi cos\phi$. When converting the integral how do you know which order they are in, I understand how you've found $R=1$ and $R=2$ and i'm assuming $\phi=0$ and $\phi=\frac{\pi}{2}$ because $z\geq0$. I am still unsure how you obtain the integral for the $\theta$. Thanks! $\endgroup$ – Ben Jones Jan 8 '18 at 11:14
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    $\begingroup$ You got it! $R^2sin\phi $ is the same Jacobian factor as of spherical coordinates (Extra information: also $r$ is similarly for Cylindrical coordinates. These two are the most important coordinates and useful specially in Electromagnetism). Since The volume has been surrounded be cone $z^2={1\over 3}(x^2+y^2)$ using coordination conversion you can obtain $R^2 cos^2(\theta)={R^2\over 3}{R^2 sin^2(\theta)}$ which yields to $tan^2\theta=3$ and for $z>0$ to $0\le \theta\le {\pi\over 3}$ $\endgroup$ – Mostafa Ayaz Jan 8 '18 at 11:24

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