Since we havex2+y2+z2≤r2,we obtain ρ=0→r,ϕ=0→π andθ=0→2π.So,we get∭Dc dV =∫02π∫0π∫0rc ρ2 sin(ϕ) dρ dϕ dθ =−c ρ33∣0r cos(ϕ)∣0πθ∣02π =−c r33( cos(π)−cos(0)(2π−0) =−c r33(−2)(2π) =43πcr3So, the correct answer is bSince \ we \ have \\ x^2+y^2+z^2\leq r^2, \\ we \ obtain\ \\ \rho=0\rightarrow r ,\\ \phi=0\rightarrow \pi\ and \\ \theta=0\rightarrow 2\pi. \\ So, we \, get\\ \iiint\limits_D c\ dV\ =\int\limits_0^{2\pi}\int\limits_0^{\pi}\int\limits_0^{r} c\ \rho^2\ sin(\phi ) \ d\rho \ d\phi\ d\theta\\ \\ \\ \qquad \qquad\ = -c\ \frac{\rho^3}{3}|_0^{r}\ cos(\phi )\large|_0^{\pi}\theta|_0^{2\pi}\\ \qquad \quad\ \ =-c\ \frac{r^3}{3}(\ cos(\pi )-cos(0)(2\pi-0)\\ \qquad \quad\ \ =-c\ \frac{r^3}{3}(-2)(2\pi)\\ \qquad \quad\ \ = \frac{4}{3} \pi c r^3\\ \text{So, the correct answer is} \ \ \ bSince we havex2+y2+z2≤r2,we obtain ρ=0→r,ϕ=0→π andθ=0→2π.So,wegetD∭c dV =0∫2π0∫π0∫rc ρ2 sin(ϕ) dρ dϕ dθ =−c 3ρ3∣0r cos(ϕ)∣0πθ∣02π =−c 3r3( cos(π)−cos(0)(2π−0) =−c 3r3(−2)(2π) =34πcr3So, the correct answer is b
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