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Use polar coordinates to find the volume of the given solid.
bounded by the paraboloids
z=12-x^(2)-y^(2) and
z=5x^(2)+5y^(2)

Use polar coordinates to find the volume of the given solid. $\newline$ bounded by the paraboloids $z=12-x^{2}-y^{2}$ and $z=5 x^{2}+5 y^{2}$

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Convert complex numbers between rectangular and polar form

Full solution

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

\newline

bounded by the paraboloids

z=12-x^{2}-y^{2}

and

z=5 x^{2}+5 y^{2}

Convert to Polar Coordinates: Convert the equations of the paraboloids to polar coordinates. $\newline$ $z = 12 - x^2 - y^2$ becomes $z = 12 - r^2$ . $\newline$ $z = 5x^2 + 5y^2$ becomes $z = 5r^2$ .
Find Intersection of Paraboloids: Find the intersection of the two paraboloids by setting their equations equal to each other. $\newline$ $12 - r^2 = 5r^2$ . $\newline$ Solve for $r^2$ . $\newline$ $r^2 = \frac{12}{6} = 2$ . $\newline$ Take the square root to find $r$ . $\newline$ $r = \sqrt{2}$ .
Set Up Volume Integral: Set up the integral to find the volume. The volume $V$ is the integral from $0$ to $2\pi$ of the integral from $0$ to $\sqrt{2}$ of $(12 - r^2) - (5r^2)$ times $r$ $dr$ $d\theta$ .
Simplify Integrand: Simplify the integrand. $\newline$ $(12 - r^2) - (5r^2) = 12 - 6r^2$ . $\newline$ Now multiply by $r$ to get the integrand for the volume. $\newline$ $12r - 6r^3$ .
Integrate with Respect to r: Integrate with respect to $r$ from $0$ to $\sqrt{2}$ . The integral of $12r$ is $6r^2$ . The integral of $-6r^3$ is $-1.5r^4$ . Evaluate from $0$ to $\sqrt{2}$ .
Plug in Limits: Plug in the limits of integration. $\newline$ For $r = \sqrt{2}$ , we have $6*(2) - 1.5*(2^2)$ . $\newline$ For $r = 0$ , we have $0$ . $\newline$ Subtract the two results to get the volume for one slice. $\newline$ $12 - 6 = 6$ .

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