Question

use polar coordinates to find the volume of the given solid. enclosed by the hyperboloid −x2 − y2 z2 = 6 and the plane z = 3

Answer 1

The volume of the solid enclosed by the hyperboloid
`\( -x^2 - y^2 + z^2 = 6 \)` and the plane ( z = 3 ) in polar coordinates is
`\( (72)/(5) \pi \)`cubic units.

Step-by-step explanation:

To find the volume of the solid, we can express the given surfaces in polar coordinates. The hyperboloid equation
`\( -x^2 - y^2 + z^2 = 6 \)` can be written as
`\( r^2 = 6\cos^2(\theta) + 6\sin^2(\theta) + 3^2 \)`, where ( r ) is the radial distance,
`\( \theta \)` is the angle in polar coordinates, and ( z = 3 ) represents the upper limit of integration.

To set up the triple integral, we need to determine the limits of integration. Since ( z ) ranges from 0 to 3,
`\( \theta \)` from 0 to
`\( 2\pi \)`, and ( r ) from 0 to
`\( √(6\cos^2(\theta) + 6\sin^2(\theta) + 9) \)`, the volume integral becomes
`\( \int_(0)^(2\pi) \int_(0)^(√(6\cos^2(\theta) + 6\sin^2(\theta) + 9)) \int_(0)^(3) r \, dz \, dr \, d\theta \).`

Solving this integral, we get
`\( (72)/(5) \pi \)` cubic units as the final answer for the volume of the given solid. This result is obtained by evaluating the triple integral with the specified limits, taking into account the polar coordinate representation of the surfaces. The calculation involves the integration of ( r ) with respect to ( z ) and
`\( \theta \),` incorporating the given constraints for the region in polar coordinates.

Answer 2

To find the volume of the solid enclosed by the hyperboloid −x² − y² + z² = 6 and the plane z = 3 using polar coordinates, we integrate the cross-sectional area times the height from the minimum radius to the maximum radius.

Step-by-step explanation:

To find the volume of the solid enclosed by the hyperboloid −x² − y² + z² = 6 and the plane z = 3 using polar coordinates, we need to set up the integral in polar form. First, we convert the equation of the hyperboloid to polar coordinates by substituting x = r cosθ and y = r sinθ. This gives us −r² + z² = 6. Rearranging the equation, we get z = √(6 + r²). Now, to find the volume, we integrate the cross-sectional area times the height from the minimum radius to the maximum radius.

The minimum radius occurs when the hyperboloid intersects with the plane z = 3. Substituting z = 3 into the equation of the hyperboloid, we get −r² + 9 = 6, which gives us r = √3. The maximum radius occurs when the hyperboloid intersects with the xy-plane, which happens when z = 0. Substituting z = 0 into the equation of the hyperboloid, we get −r² = 6, which gives us r = √(-6).

Therefore, the volume of the solid can be calculated as follows:

V = ∫[0 to 2π] ∫[√3 to √(-6)] √(6 + r²) r dr dθ

Integrating this expression will give us the volume of the given solid in polar coordinates.