Answer to Question #174760 in Calculus for Muhammad Ahmed

Question #174760

With the help of triple integrals, find the volume of the sphere ρ = 2 in a) Rectangular coordianates b) Cylinderical coordiantes c) Spherical coordinates 


1
Expert's answer
2021-04-29T17:55:56-0400

With the help of triple integrals, find the volume of the sphere ρ = 2 in a) Rectangular coordianates b) Cylinderical coordiantes c) Spherical coordinates 

Solution:

a) Rectangular coordinates:


"x^2+y^2+z^2=4"

"V=\\displaystyle\\int_{-2}^2dx\\int_{-\\sqrt{4-x^2}}^{\\sqrt{4-x^2}}dy\\int_{-\\sqrt{4-x^2-y^2}}^{\\sqrt{4-x^2-y^2}}dz=" "8\\displaystyle\\int_{0}^2dx\\int_{0}^{\\sqrt{4-x^2}}dy\\int_{0}^{\\sqrt{4-x^2-y^2}}dz=" "8\\displaystyle\\int_{0}^2dx\\int_{0}^{\\sqrt{4-x^2}}\\sqrt{4-x^2-y^2}dy"

Let's apply the change of variables:

"y=\\sqrt{4-x^2}\\cdot\\sin{t}"

"dy=\\sqrt{4-x^2}\\cdot\\cos{t}\\cdot dt"

"y=0 \\implies t=0"

"y=\\sqrt{4-x^2}\\implies t=\\frac{\\pi}{2}"

"V=8\\displaystyle\\int_{0}^2dx\\int_{0}^{\\pi\/2}\\sqrt{4-x^2-(4-x^2)\\sin^2{t}}\\cdot" "\\sqrt{4-x^2}\\cos{t}dt="

"8\\displaystyle\\int_{0}^2dx\\int_{0}^{\\pi\/2}(4-x^2)\\cos^2{t}dt=" "8\\displaystyle\\int_{0}^2dx\\int_{0}^{\\pi\/2}(4-x^2)\\frac{(1+\\cos{2t})}{2}dt=" "8\\displaystyle\\int_{0}^2dx(4-x^2)\\frac{(t+\\frac{\\sin{2t}}{2})}{2}|_0^{\\pi\/2}=" "2\\pi\\displaystyle\\int_{0}^2(4-x^2)dx=2\\pi(4x-x^3\/3)|_0^2=\\frac{32\\pi}{3}"

b) Cylinderical coordiantes:

"x=\\rho\\cos{\\varphi}"

"y=\\rho\\sin{\\varphi}"

"z=z"

"J=\\rho"

The equation of the sphere:

"\\rho^2+z^2=4"

"V=\\displaystyle\\int_0^{2\\pi}d\\varphi\\int_0^2d\\rho\\int_{-\\sqrt{4-\\rho^2}}^{\\sqrt{4-\\rho^2}}\\rho dz=" "2\\displaystyle\\int_0^{2\\pi}d\\varphi\\int_0^2d\\rho\\int_0^{\\sqrt{4-\\rho^2}}\\rho dz=" "4\\pi\\displaystyle\\int_0^2\\sqrt{4-\\rho^2}\\rho d\\rho=" "2\\pi\\displaystyle\\int_0^2\\sqrt{4-\\rho^2} d\\rho^2=-\\frac{4\\pi}{3}\\displaystyle(4-\\rho^2)^{3\/2}|_0^2=\\frac{32\\pi}{3}"

c) Spherical coordinates 

"x=r\\sin{\\theta}\\cos{\\varphi}"

"y=r\\sin{\\theta}\\sin{\\varphi}"

"z=r\\cos{\\theta}"

"J=r^2\\sin{\\theta}"

"V=\\displaystyle\\int_0^{2\\pi}d\\varphi\\int_0^{\\pi}d\\theta\\int_0^2r^2\\sin{\\theta}dr=" "2\\pi\\displaystyle\\int_0^{\\pi}d\\theta\\sin{\\theta}\\frac{r^3}{3}|_0^2=" "\\displaystyle\\frac{16\\pi}{3}\\int_0^{\\pi}\\sin{\\theta}d\\theta=" "-\\displaystyle\\frac{16\\pi}{3}\\cos{\\theta}|_0^{\\pi}=\\frac{32\\pi}{3}"

Answer: "V=\\displaystyle\\frac{32\\pi}{3}."

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