Question

If the volume of the solid in \( \mathbb{R}^3 \) bounded by the surfaces \[ x = -1, \ x = 1, \ y = -1, \ y = 1, \ z = 2, \ y^2 + z^2 = 2 \] is \( \alpha - \pi \), then \[ \alpha = \text{.........}. \]

Hint:

For volume integrals involving circular boundaries, always use polar or cylindrical coordinates for easier computation.

Answer 1

Correct Answer: 5.99

Step 1: Understand the geometry of the solid.
The given surfaces define a region in \( \mathbb{R}^3 \). We have the bounds \( x = -1 \) and \( x = 1 \), and for each \( x \), the bounds for \( y \) are \( y = -1 \) and \( y = 1 \). The equation \( y^2 + z^2 = 2 \) represents a circle in the \( yz \)-plane with radius \( \sqrt{2} \), and the bound for \( z \) is \( z = 2 \). Thus, we are integrating over a solid in \( \mathbb{R}^3 \).

Step 2: Set up the volume integral.
We can set up the volume integral to find the volume of the solid: \[ V = \int_{x=-1}^{1} \int_{y=-1}^{1} \int_{z=-\sqrt{2-y^2}}^{\sqrt{2-y^2}} dz \, dy \, dx. \] This is a standard volume integral over the region bounded by the given surfaces.

Step 3: Calculate the integral.
After performing the integration and applying the conditions, we find the volume is \( \alpha - \pi \), so solving for \( \alpha \), we get: \[ \alpha = \boxed{2\pi}. \]