Answer to Question #106868 in Analytic Geometry for khushi

Question #106868
a) Express the following surfaces in spherical coordinates
i) xz = 3
ii)x^2+y^2-z^2=1
1
Expert's answer
2020-03-30T09:21:51-0400

"1)\\ xz=3"

"x=r\\sin{\\theta}\\cos{\\phi} \\\\\nz=r\\cos{\\theta}\\\\\nxz= r\\sin{\\theta}\\cos{\\phi}\\ r\\cos{\\theta}=r^2\\sin{\\theta}\\cos{\\theta}\\cos{\\phi}=\\frac{1}{2}r^2\\sin{2\\theta}\\cos{\\phi}=3\\\\\nr^2\\sin{2\\theta}\\cos{\\phi}=6, \\ r>0,\\ \\theta\\in[0,\\pi], \\ \\phi\\in[0,2\\pi]"


"2)\\ x^2+y^2-z^2=1\\\\\nx=r\\sin{\\theta}\\cos{\\phi} \\\\\ny=r\\sin{\\theta}\\sin{\\phi} \\\\\nz=r\\cos{\\theta}\\\\\nx^2+y^2-z^2=r^2\\sin^2{\\theta}\\cos^2{\\phi}+r^2\\sin^2{\\theta}\\sin^2{\\phi}-r^2\\cos^2{\\theta}=\\\\\n=r^2\\sin^2{\\theta}(\\cos^2{\\phi}+\\sin^2{\\phi})-r^2\\cos^2{\\theta}=r^2\\sin^2{\\theta}-r^2\\cos^2{\\theta}=\\\\\n=r^2(\\sin^2{\\theta}-\\cos^2{\\theta})=-r^2\\cos{2\\theta}=1\\\\\nr^2\\cos{2\\theta}=-1,\\ r>0,\\ \\theta\\in[0,\\pi], \\ \\phi\\in[0,2\\pi]"


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