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Use polar coordinates to find the volume of the given solid.
above the cone
z=sqrt(x^(2)+y^(2)) and below the sphere
x^(2)+y^(2)+z^(2)=81

Use polar coordinates to find the volume of the given solid. $\newline$ above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the sphere $x^{2}+y^{2}+z^{2}=81$

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Convert complex numbers between rectangular and polar form

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Q. Use polar coordinates to find the volume of the given solid.

\newline

above the cone

z=\sqrt{x^{2}+y^{2}}

and below the sphere

x^{2}+y^{2}+z^{2}=81

Write Equations: First, let's write the equations of the cone and the sphere in polar coordinates. For the cone, we have $z=r$ (since $z=\sqrt{x^2+y^2}$ and in polar coordinates, $x=r\cos(\theta)$ , $y=r\sin(\theta)$ , so $x^2+y^2=r^2$ ). For the sphere, we have $r^2+z^2=81$ .
Set Up Integral: Now, we'll set up the integral to find the volume. We need to integrate in the $r$ and $\theta$ directions, as well as the $z$ direction. The limits for $\theta$ are from $0$ to $2\pi$ , since it's a full rotation around the $z$ -axis. The limits for $r$ will be from $0$ to the intersection of the cone and sphere, and the limits for $z$ will be from the cone ( $\theta$ $0$ ) to the sphere ( $\theta$ $1$ ).
Find Intersection: To find the intersection of the cone and sphere, we set their equations equal: $r=\sqrt{81-r^2}$ . Squaring both sides, we get $r^2=81-r^2$ , which simplifies to $2r^2=81$ , so $r^2=40.5$ , and $r=\sqrt{40.5}$ .

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