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Evaluate the integral by changing to spherical coordinates.

$ \large \int_{-4}^4 \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^4 \sqrt{x^2+y^2+z^2} \ dzdydx$

Answer:

$ x=\rho \sin \phi \cos \theta, \\ y=\rho \sin \phi \sin \theta, \\ z=\rho \cos \phi , \ $

So $ \ dzdydx=\rho^2 \sin \phi \ \ d \rho d \phi d \theta \ $

The ranges will be $ \ 0 \leq \theta \leq 2 \pi , \ \ 0 \leq \rho \leq 4 \ $

Now to determine the ranges of $ \ \phi \ $

$ z=\sqrt{x^2+y^2} \ $ gives $ \ \phi=\frac{\pi}{4} \ $,

$ z=4 \ $ gives $ \ \rho \cos \phi=4 \ \Rightarrow \phi=\cos^{-1} (\frac{4}{\rho} ) \ $

Thus the new integral should be

$ \int_{0}^{2 \pi} \int_{\frac{\pi}{4}}^{\cos^{-1}(\frac{4}{\rho})} \int_{0}^{4} \rho \cdot \rho^2 \sin \phi \ d \rho d \phi d \theta \ $

I need confirmation of my work.

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  • $\begingroup$ so what would be the setting ? $\endgroup$ – M. A. SARKAR Apr 6 '18 at 0:50
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Your contraints on your region are:

$z \le 4\\ z^2 \ge x^2 + y^2 $

and $x^2 + y^2 \le 16$

The region is a cone.

Make our substitutions

$\rho\cos \phi \le 4\\ \rho^2\cos^2\phi \ge \rho^2\sin^2\phi\\ \rho^2\sin^2\phi \le 4$

$ \rho \le 4\sec\phi\\ \tan \phi \le 1\\ \phi \le \frac {\pi}{4}$

The last constraint doesn't turn out to be relevant.

Almost always, we want $\rho$ in terms of $\theta$ and $\phi$ and not the other way around.

$\int_0^{2\pi}\int_0^{\frac {\pi}{4}}\int_0^{4\sec\phi} \rho^3\sin\phi \ d\rho\ d\phi\ d\theta$

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