Use a triple integral to determine the volume of the region bounded by z =
p
x
2 + y
2 and z = x
2 + y
2
In 1st octant
To convert from cylindrical to rectangular coordinates, we use the equations
"z=\\sqrt{x^2+y^2}, z=\\sqrt{(r\\cos \\theta)^2+(r\\sin \\theta)^2}=r"
"z=x^2+y^2, z=(r\\cos \\theta)^2+(r\\sin \\theta)^2=r^2"
"r=r^2, r_1=0, r_2=1"
"V=\\displaystyle\\int_{0}^{\\pi\/2}d\\theta\\displaystyle\\int_{0}^{1}rdr\\displaystyle\\int_{r^2}^{r}dz"
"=\\displaystyle\\int_{0}^{\\pi\/2}d\\theta\\displaystyle\\int_{0}^{1}(r^2-r^3)dr"
"=\\dfrac{\\pi}{2}(\\dfrac{1}{3}-\\dfrac{1}{4})=\\dfrac{\\pi}{24}({units}^2)"
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