- #1
tintin2006
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I want to check if I'm doing this problem correctly.
Region bounded by x^2+y^2=4 and bounded by the surfaces z = 0, and z=\sqrt{9-x^2-y^2}.
Set up triple integrals which represent the volume of the solid using spherical coordinates.
\int\int\int_{V}\rho^2sin\phi \; d\rho d\phi d\theta
The shape is a cylinder with a round top.
It seems that I have to break this into two integrals:
1:
From x^2+y^2=4 and z=\sqrt{9-x^2-y^2}:
z=\sqrt{9-4}=\sqrt{5}
Since z=\rho cos\phi, \rho cos\phi = \sqrt{5} \Rightarrow \rho = \frac{\sqrt{5}}{cos\phi}
2:
From z=\sqrt{9-x^2-y^2}, 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:
sin\phi = \frac{2}{3}
\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
Thank you.
Homework Statement
Region bounded by x^2+y^2=4 and bounded by the surfaces z = 0, and z=\sqrt{9-x^2-y^2}.
Set up triple integrals which represent the volume of the solid using spherical coordinates.
Homework Equations
\int\int\int_{V}\rho^2sin\phi \; d\rho d\phi d\theta
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 x^2+y^2=4 and z=\sqrt{9-x^2-y^2}:
z=\sqrt{9-4}=\sqrt{5}
Since z=\rho cos\phi, \rho cos\phi = \sqrt{5} \Rightarrow \rho = \frac{\sqrt{5}}{cos\phi}
2:
From z=\sqrt{9-x^2-y^2}, 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:
sin\phi = \frac{2}{3}
\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
Thank you.
