Question

Let \( \Omega \) be the bounded region in \( \mathbb{R}^3 \) lying in the first octant \( (x \geq 0, y \geq 0, z \geq 0) \), and bounded by the surfaces \( z = x^2 + y^2 \), \( z = 4 \), \( x = 0 \) and \( y = 0 \). Then, the volume of \( \Omega \) is equal to:

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( 3\pi \)
• D. \( 4\pi \)

Hint:

When dealing with volume integrals, carefully analyze the region and set up the limits based on the given bounds. Use symmetry when possible to simplify the computation.

Answer 1

The Correct Option is B

Step 1: Set up the limits for integration.
We are given the region \( \Omega \) bounded by \( z = x^2 + y^2 \), \( z = 4 \), \( x = 0 \), and \( y = 0 \). To find the volume, set up the triple integral: \[ V = \int_{0}^{2} \int_{0}^{\sqrt{4 - x^2}} \int_{x^2 + y^2}^{4} dz \, dy \, dx. \] Step 2: Integrate with respect to \( z \).
The integral becomes: \[ V = \int_{0}^{2} \int_{0}^{\sqrt{4 - x^2}} \left[ z \right]_{z = x^2 + y^2}^{z = 4} dy \, dx = \int_{0}^{2} \int_{0}^{\sqrt{4 - x^2}} (4 - x^2 - y^2) \, dy \, dx. \] Step 3: Integrate with respect to \( y \).
Integrating with respect to \( y \) gives: \[ V = \int_{0}^{2} \left[ (4 - x^2) y - \frac{y^3}{3} \right]_{0}^{\sqrt{4 - x^2}} dx. \] After substituting the limits: \[ V = \int_{0}^{2} \left( (4 - x^2)\sqrt{4 - x^2} - \frac{(4 - x^2)^{3/2}}{3} \right) dx. \] Step 4: Final evaluation of the integral.
The exact volume, after evaluating this integral, gives \( 2\pi \). Final Answer: \[ \boxed{2\pi}. \]