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Chapter 12: Problem 12

Use cylindrical coordinates to find the volume of the solid. Solid inside the sphere \(x^{2}+y^{2}+z^{2}=4\) and above the upper nappe of the cone \(z^{2}=x^{2}+y^{2}\)

Short Answer

Expert verified
The volume of the solid inside the sphere and above the upper nappe of the cone is \(\frac{\pi}{6}\) cubic units.

Step by step solution

01

Convert the equations into cylindrical coordinates

The equation of the sphere is \(x^{2}+y^{2}+z^{2}=4\) and the equation of the cone is \(z^{2}=x^{2}+y^{2}\). We converted this to cylindrical coordinates using \(x = r\cos\theta, y = r\sin\theta, z = z\). For the sphere the equation becomes \(r^{2} + z^{2} = 4\), and for the cone the equation becomes \(z^{2} = r^{2}\). In other words, \(z = r\) for the upper nappe of the cone, which is implied by the non-negativity of z in cylindrical coordinates.
02

Set up the limits of integration

To find the region of the solid, we need to set the bounds for \(r\), \(z\), and \(\theta\). The solid is entirely within the x-y plane, so the bounds for \(\theta\) are \([0, 2\pi]\). The intersection of the sphere and the cone gives us the bounds for \(r\) and \(z\). Solving the equation \(r^{2} + r^{2} = 4\) (derived from sphere and cone intersection) gives us \(r = 1\), therefore, the bounds for \(r\) and \(z\) are respectively \([0,1]\) and \([r, \sqrt{4-r^2}]\).
03

Integrate to find the volume

The volume is given by, \(V = \int \int \int r \, dz \, dr \, d\theta\). Now perform the integration, \[ V = \int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta \]. The practicalities of this calculation depends on your knowledge of calculus. After performing the calculation we find the volume \(V = \frac{\pi}{6}\].

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