I want to check if I'm doing this problem correctly.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Region bounded by [itex]x^2+y^2=4[/itex] and bounded by the surfaces z = 0, and [itex]z=\sqrt{9-x^2-y^2}[/itex].

Set up triple integrals which represent the volume of the solid using spherical coordinates.

2. Relevant equations

[itex]\int\int\int_{V}\rho^2sin\phi \; d\rho d\phi d\theta[/itex]

3. The attempt at a solution

The shape is a cylinder with a round top.

It seems that I have to break this into two integrals:

1:

From [itex]x^2+y^2=4[/itex] and [itex]z=\sqrt{9-x^2-y^2}[/itex]:

[itex]z=\sqrt{9-4}=\sqrt{5}[/itex]

Since [itex]z=\rho cos\phi[/itex], [itex]\rho cos\phi = \sqrt{5} \Rightarrow \rho = \frac{\sqrt{5}}{cos\phi}[/itex]

2:

From [itex]z=\sqrt{9-x^2-y^2}[/itex], it is a sphere of radius 3 so ρ=3

The sphere and cylinder meets at point (y,z)=(2, √5) and the radius makes an angle with the point:

[itex]sin\phi = \frac{2}{3}[/itex]

[itex]\int_{0}^{2\pi}\int_{0}^{arcsin(\frac{2}{3})}\int_{0}^{3}\rho^2sin\phi \; d\rho d\phi d\theta + \int_{0}^{2\pi}\int_{arcsin(\frac{2}{3})}^{\frac{ \pi }{2}}\int_{0}^{\frac{\sqrt{5}}{cos\phi}}\rho^2 sin \phi \; d\rho d\phi d\theta[/itex]

Thank you.

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# Triple integral in spherical coordinates

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