"\\text{As we have}\\\\\nx^{2}+y^{2}+z^{2} \\leq r^{2}\\\\\nwe\\ get\\\\\n\\rho=0 \\rightarrow r ,\\\\\n\\phi=0 \\rightarrow \\pi \\text { and } \\\\\n\\theta=0 \\rightarrow 2 \\pi \\\\\n\\text { So, we obtain} \\\\\n\\iiint\\limits_{D} c d V=\\int_{0}^{2 \\pi} \\int_{0}^{\\pi} \\int_{0}^{r} c \\rho^{2} \\sin (\\phi) d \\rho d \\phi d \\theta \\\\\n=-\\left.\\left.\\left.c \\frac{\\rho^{3}}{3}\\right|_{0} ^{r} \\cos (\\phi)\\right|_{0} ^{\\pi} \\theta\\right|_{0} ^{2 \\pi}\\\\\n=-c \\frac{r^{3}}{3}(\\cos (\\pi)-\\cos (0)(2 \\pi-0))\\\\\n=-c \\frac{r^{3}}{3}(-2)(2 \\pi) \\\\\n=\\frac{4}{3} \\pi c r^{3}\\\\\n\\text{Then, the correct answer is (b)}"
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