Question:

Evaluate

\(\iiint_D x\sqrt{1+x^2+y^2+z^2}dxdydz\)

The domain is \(D=\{(x,y,z) | x\geq 0, y\geq 0, x^2+y^2+z^2\leq 1 \}\).

Answer:

Use the polar coordinate.

\[  x = r\sin\theta \cos\phi \]

\[  y = r\sin\theta \sin\phi  \]

\[  z = r\cos\theta  \]

Therefore, the Jacobian becomes \(J=r^2\sin\theta\). Then, substitute them into the original integral as follows:

\[  I = \iiint_D x\sqrt{1+x^2+y^2+z^2}dxdydz　　　\]

\[  =  {\scriptstyle \iiint_D r^3\sin^2\theta \cos\phi\sqrt{1+(r\sin\theta \cos\phi)^2+(r\sin\theta \sin\phi)^2+(r\cos\theta)^2}dr d\theta d\phi  }      \]

\[  = {\scriptstyle \iiint_D r^3\sin^2\theta \cos\phi\sqrt{1+r^2\sin^2\theta(\cos^2\phi + \sin^2\phi)+r^2\cos^2\theta}dr d\theta d\phi  }      \]

\[  = \iiint_D r^3\sin^2\theta \cos\phi\sqrt{1+r^2\sin^2\theta+r^2\cos^2\theta}dr d\theta d\phi   \]

\[  = \iiint_D r^3\sin^2\theta \cos\phi\sqrt{1+r^2}dr d\theta d\phi   \]

\[  = \int^1_0 r^3\sqrt{1+r^2}dr \int^{\pi}_0 \sin^2\theta d\theta \int^{\pi/2}_0\cos\phi  d\phi   \]

The domain, \(D\), explains that \(0\leq r \leq 1\), \(0\leq \theta \leq \pi\), and \(0\leq \phi \leq \pi/2\). Let us calculate each part of the integral.

\[  I_1 = \int^1_0 r^3\sqrt{1+r^2} dr  \]

\[      = \int^2_1 (t-1)\sqrt{t} dt \quad {\scriptstyle (\mathtt{Replaced \ with} \ 1+r^2=t.)}  \]

\[      = \frac{1}{2}\left[\frac{2}{5}t^{5/2}-\frac{2}{3}t^{3/2}\right]^2_1  \]

\[      = \frac{2}{15}(\sqrt{2}+1)  \]

\[  I_2 = \int^{\pi}_0 \sin^2\theta d\theta   \]

\[      = \int^{\pi}_0 \frac{1-\cos 2\theta}{2} d\theta   \]

\[      = \int^{\pi}_0 \frac{1}{2}-\frac{\cos 2\theta}{2} d\theta   \]

\[      = \left[ \frac{\theta}{2}-\frac{\sin 2\theta}{4} \right]^{\pi}_0    \]

\[      = \frac{\pi}{2}   \]

\[  I_3 = \int^{\pi/2}_0 \cos\phi d\phi   \]

\[      = \left[ \sin\phi \right]^{\pi/2}_0   \]

\[      = 1  \]

Therefore,

\[  I = I_1\cdot I_2\cdot I_3 = \frac{2}{15}(\sqrt{2}+1)\cdot \frac{\pi}{2} \cdot 1 = \frac{\pi}{15}(\sqrt{2}+1)  \]