Answer to Question #119473 in Calculus for Olivia

Question #119473
Let c,r be constants, and D={(x,y,z):x^2+y^2+z^2≤r^2}. The answer to ∭D cdV

is
Select one:
a. (π^2cr^3)/3


b. (4πcr^3)/3

c. 4πcr^3

d. (πcr^4)/3


e. (4πcr^2)/2

f. (4r^3)/3


g. (πcr^3)/3
1
Expert's answer
2020-06-07T16:19:00-0400

"\\text { The easiest way to do this is to make a switch to spherical coordinates.} \\\\[1 em] \n\\text { There }\\rho^{2}=x^{2}+y^{2}+z^{2} \\text { \\, and\\,} d x d y d z=\\rho^{2} \\sin \\phi d \\rho d\\theta d \\varphi. \\\\[1 em]\n\\iiint_{D} c dv=\\iiint_{D} c d x d y d z =\\iiint_{D} c \\cdot \\rho^{2} \\sin \\varphi d \\rho d\\theta d \\varphi\n\\\\[1 em]\n\\text { Now we are integrating over a region} D.\\\\[1 em]\n \\text{What is} D\\text{? It is a sphere of radius r centered at the origin.}\\\\[1 em]\n\\begin{array}{l}\n\\text { So } 0 \\leq \\rho \\leq r,0 \\leq \\theta \\leq 2 \\pi \\text { , and } 0 \\leq \\varphi \\leq \\pi \\\\[1 em]\n\\qquad \\iiint_{D}c \\rho^{2} \\sin \\varphi d \\rho d \\theta d \\varphi=\\int_{0}^{\\pi} \\int_{0}^{2 \\pi} \\int_{0}^{r} c \\rho^{2} \\sin \\varphi d \\rho d \\theta d \\varphi\n\\end{array}\n\\\\[1 em]\n\n\\\\[1 em]\n\n \\begin{aligned}\nc\\int_{0}^{2 \\pi} d \\theta \\int_{0}^{\\pi} \\sin \\varphi d \\varphi \\int_{0}^{r} \\rho^{2} d \\rho\n&= c (\\theta\\bigg| _{0}^{2 \\pi} )(-\\cos\\varphi\\bigg| _{0}^{\\pi} )(\\frac{ \\rho^{3} }{3}\\bigg| _{0}^{r}) \\\\\n&=c(2 \\pi)(-(-1-1))(\\frac{ r^{3} }{3})\\\\\n&=\\frac{4\\pi cr^3}{3}\n\\end{aligned}\\\\\n\\text{The answer is } \\fcolorbox{red}{aqua}{b}"


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