Question

The value of \( \iiint_E \frac{dx\,dy\,dz}{x^2+y^2+z^2} \), where E: \( x^2+y^2+z^2 \le a^2 \), is

• A. \( \pi a \)
• B. \( 2\pi a \)
• C. \( 4\pi a \)
• D. \( 8\pi a \)

Hint:

Anytime you see a triple integral over a spherical region (sphere, hemisphere, etc.) and the integrand involves \(x^2+y^2+z^2\), immediately switch to spherical coordinates. The Jacobian \( \rho^2 \sin\phi \) will often simplify the expression.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem asks for the evaluation of a triple integral over a spherical region. The integrand and the region of integration are both symmetric with respect to the origin and involve the term \( x^2+y^2+z^2 \), which strongly suggests using spherical coordinates.

Step 2: Key Formula or Approach:
The transformation to spherical coordinates is: - \( x = \rho \sin\phi \cos\theta \) - \( y = \rho \sin\phi \sin\theta \) - \( z = \rho \cos\phi \) The term \( x^2+y^2+z^2 = \rho^2 \). The volume element is \( dV = dx\,dy\,dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \). The region \( E: x^2+y^2+z^2 \le a^2 \) corresponds to a solid sphere of radius \(a\), so the limits are: - \( 0 \le \rho \le a \) - \( 0 \le \phi \le \pi \) - \( 0 \le \theta \le 2\pi \)

Step 3: Detailed Explanation:
Substitute the spherical coordinates into the integral: \[ \iiint_E \frac{1}{x^2+y^2+z^2} dx\,dy\,dz = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \frac{1}{\rho^2} (\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta) \] The \( \rho^2 \) terms cancel out, simplifying the integral significantly: \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \sin\phi \, d\rho \, d\phi \, d\theta \] The integrand now only depends on \( \phi \), so we can separate the integrals: \[ \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{\pi} \sin\phi \, d\phi \right) \left( \int_{0}^{a} d\rho \right) \] Evaluate each integral separately: - \( \int_{0}^{2\pi} d\theta = [\theta]_{0}^{2\pi} = 2\pi \) - \( \int_{0}^{\pi} \sin\phi \, d\phi = [-\cos\phi]_{0}^{\pi} = -(\cos\pi - \cos0) = -(-1 - 1) = 2 \) - \( \int_{0}^{a} d\rho = [\rho]_{0}^{a} = a \) Multiply the results: \[ \text{Value} = (2\pi) \times (2) \times (a) = 4\pi a \]
Step 4: Final Answer:
The value of the integral is \( 4\pi a \).