Question

Answer 1

The integral bounded by region \(D\) is given by

\(H=\iiint_{D} 1 d V\)

where the region \(\mathrm{D}\) is described by

\(x^{2}+y^{2}+z^{2} \leq \sqrt{x^{2}+y^{2}}\)

\(z \geq 0\)

if we convert the region into the spherical coordinates as

\(x=\rho \sin (\phi) \cos (\theta)\)

\(y=\rho \sin (\phi) \sin (\theta)\)

\(z=\rho \cos (\phi)\)

then the inequality becomes

\(x^{2}+y^{2}+z^{2} \leq \sqrt{x^{2}+y^{2}}\)

\(\rho^{2} \leq \sqrt{\rho^{2} \sin ^{2}(\phi)}\)

\(\rho \leq \sin (\phi)\)

the region \(\mathrm{D}\) is described as

\(0 \leq \rho \leq \sin (\phi)\)

\(0 \leq \theta \leq 2 \pi\)

$$ \begin{aligned} &\text { then the integral becomes }\\ &H=\iiint_{D} 1 d V\\ &H=\iiint_{D} \rho^{2} \sin (\phi) d \rho d \phi d \theta\\ &H=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin (\phi)} \rho^{2} \sin (\phi) d \rho d \phi d \theta\\ &H=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}}\left[\frac{\rho^{3}}{3}\right]_{0}^{\sin (\phi)} \sin (\phi) d \phi d \theta\\ &H=\frac{1}{3} \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \sin ^{4}(\phi) d \phi d \theta\\ &H=\frac{1}{3} \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}}\left(\frac{3-4 \cos (2 \phi)+\cos (4 \phi)}{8}\right) d \phi d \theta\\ &H=\frac{1}{24} \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}}(3-4 \cos (2 \phi)+\cos (4 \phi)) d \phi d \theta\\ &H=\frac{1}{24} \int_{0}^{2 \pi}\left[3 \phi-2 \sin (2 \phi)+\frac{\sin (4 \phi)}{4}\right]_{0}^{\frac{\pi}{2}} d \theta\\ &H=\frac{1}{24}\left(\frac{3 \pi}{2}\right) \int_{0}^{2 \pi} d \theta\\ &H=\frac{1}{24}\left(\frac{3 \pi}{2}\right)[\theta]_{0}^{2 \pi}\\ &H=\frac{1}{24}\left(\frac{3 \pi}{2}\right)(2 \pi)\\ &H=\frac{1}{8} \pi^{2}\\ &H=\frac{\pi^{2}}{2} \end{aligned} $$