Use a double integral in polar coordinates to find the volume of the solid in the first octant bounded by the graphs of the equations given below.

Question

Quizplus · Accepted Answer

The solid is bounded by the graphs of the equations:
$z=4-x^2-y^2$ and $z=x^2+y^2$
Converting to polar coordinates, we have:
$x=r\cos	heta$ and $y=r\sin	heta$
The equations become:
$z=4-r^2$ and $z=r^2$
To find the limits of integration, note that in the first octant, $0\leq	heta\leq\frac{\pi}{2}$ and $0\leq r\leq2$.
Thus, the volume of the solid is given by the double integral:
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}(4-r^2-r^2)rdrd	heta$
Simplifying and evaluating this integral gives the answer B.

Use a Double Integral in Polar Coordinates to Find the Volume