Question

Let \[ B = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 \le 1 \} \] and define \[ u(x,y,z) = \sin\!\left(\pi(1 - x^2 - y^2 - z^2)^2\right) \] for \((x,y,z) \in B.\) Then the value of \[ \iiint_B \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \right) \, dx\,dy\,dz \] is _________.

Hint:

If a smooth function \(u\) vanishes on the boundary of a region, then \(\displaystyle \iiint \nabla^2 u = 0\) by the divergence theorem.

Answer 1

Correct Answer: 0

Step 1: Apply divergence theorem.
Note that \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}. \] By the divergence theorem, \[ \iiint_B \nabla^2 u \, dV = \iint_{\partial B} \nabla u \cdot \mathbf{n} \, dS. \]
Step 2: Evaluate on the boundary.
On the boundary \(\partial B\), \(x^2 + y^2 + z^2 = 1.\) Then \(1 - x^2 - y^2 - z^2 = 0 \Rightarrow u = \sin(0) = 0.\) Hence, \(u\) is constant on the boundary, and \(\nabla u = 0\) there. Thus, the surface integral equals 0.
Step 3: Conclusion.
\[ \iiint_B \nabla^2 u \, dV = 0. \]