Triple integral

Question:
Evaluate
$\iiint_D x\sqrt{1+x^2+y^2+z^2}dxdydz$
The domain is $D=\{(x,y,z) | x\geqq 0, y\geqq 0, x^2+y^2+z^2\leqq 1 \}$.

Answer:
Use the polar coordinate.
\begin{eqnarray*}
x &=& r\sin\theta \cos\phi \\
y &=& r\sin\theta \sin\phi  \\
z &=& r\cos\theta
\end{eqnarray*}
Then, the Jacobian becomes $J=r^2\sin\theta$. Now substitute them into the original integral.
\begin{eqnarray}
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   \\
\end{eqnarray}
The domain, $D$, explains that $0\leqq r \leqq 1$, $0\leqq \theta \leqq \pi$, and $0\leqq \phi \leqq \pi/2$. Let us calculate each part of the integral.
\begin{eqnarray*}
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
\end{eqnarray*}
Therefore,
\[
I = I_1\cdot I_2\cdot I_3 = \frac{2}{15}(\sqrt{2}+1)\cdot \frac{\pi}{2} \cdot 1 = \frac{\pi}{15}(\sqrt{2}+1)
\]

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