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(c) The solid is above the cone \( z=\sqrt{x^{2}+y^{2}} \) and lies between the spheres \( x^{2}+y^{2}+z^{2}=4 \) and \( x^{2}+y^{2}+z^{2}=9 \)

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Final Answer:

The region described is a solid bounded by the cone
\( z=\sqrt{x^(2)+y^(2)} \) from below and the spheres
\( x^(2)+y^(2)+z^(2)=4 \) and \( x^(2)+y^(2)+z^(2)=9 \) from above.

Step-by-step explanation:

To understand this solid, let's break down the given information. The cone
\( z=\sqrt{x^(2)+y^(2)} \) represents a cone extending infinitely upwards from the xy-plane. The two spheres,
\( x^(2)+y^(2)+z^(2)=4 \) and
\( x^(2)+y^(2)+z^(2)=9 \), denote spheres with radii 2 and 3, respectively, centered at the origin.

Now, considering the constraints provided, the solid is above the cone, meaning it exists in the region where
\( z > \sqrt{x^(2)+y^(2)} \).Additionally, it lies between the spheres, indicating that the solid occupies the space between the two spheres.

In mathematical terms, the solid is defined by the inequalities
\( \sqrt{x^(2)+y^(2)} < z \) and \( 2^(2) \leq x^(2)+y^(2)+z^(2) \leq 3^(2) \). These conditions ensure that the solid is both above the cone and confined within the space between the spheres. This geometric configuration creates a unique region in three-dimensional space that satisfies all given constraints.

User Hrant Khachatrian
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