Respuesta :
The solid is a Cone.
Given the inequalities:
[tex]0\leq[/tex] θ [tex]\leq[/tex][tex]\frac{\pi }{2}[/tex] : [tex]r\leq z\leq 4[/tex]
These are given in cylindrical co-ordinates. We convert it to the cartesian co-ordinates to make it easier to understand.
Here the first inequality simply means that [tex]x\geq 0[/tex] and [tex]y\geq 0[/tex].
We have [tex]r=\sqrt{x^2+y^2}[/tex] .
Hence the second inequality implies that [tex]\sqrt{x^2+y^2} \leq z\leq 4[/tex]
This can be separated into three inequalities,
[tex]\sqrt{x^2+y^2}\leq 4[/tex] ; [tex]\sqrt{x^2+y^2} \leq z[/tex] ; [tex]z\leq 4[/tex]
The first inequality gives [tex]x^2+y^2\leq 4^2[/tex] which is the inside of a circle of radius 4.
The second inequality gives [tex]x^2+y^2\leq z^2[/tex]. This is the equation of a Cone given in cartesian co-ordinates.
The third inequality simply implies that the height of the solid(Cone) is [tex]\leq[/tex] 4.
Thus the given inequalities describe a cone with base radius [tex]\leq[/tex] 4 and height [tex]\leq[/tex] 4.
Learn more about describing solids from inequalities at https://brainly.com/question/1550790
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