Question

Let \( \Omega \) be the solid bounded by the planes \( z = 0 \), \( y = 0 \), \( x = \frac{1}{2} \), \( 2y = x \) and \( 2x + y + z = 4 \). If \( V \) is the volume of \( \Omega \), then the value of \( 64V \) is equal to .......... (rounded off to two decimal places).

Hint:

When finding the volume of a solid, express the bounds carefully and use triple integrals to compute the volume.

Answer 1

Step 1: Set up the bounds for integration.
The solid is bounded by the planes \( z = 0 \), \( y = 0 \), \( x = \frac{1}{2} \), \( 2y = x \), and \( 2x + y + z = 4 \). From the equation \( 2x + y + z = 4 \), solve for \( z \): \[ z = 4 - 2x - y. \] Thus, the volume is given by the triple integral: \[ V = \int_0^{1/2} \int_0^x \int_0^{4 - 2x - y} dz \, dy \, dx. \] Step 2: Perform the integration.
First, integrate with respect to \( z \): \[ \int_0^{4 - 2x - y} dz = 4 - 2x - y. \] Now, integrate with respect to \( y \): \[ \int_0^x (4 - 2x - y) \, dy = (4 - 2x) x - \frac{x^2}{2} = 4x - 2x^2 - \frac{x^2}{2}. \] Finally, integrate with respect to \( x \): \[ \int_0^{1/2} \left( 4x - 2x^2 - \frac{x^2}{2} \right) \, dx = \frac{4x^2}{2} - \frac{2x^3}{3} - \frac{x^3}{6} \bigg|_0^{1/2}. \] After calculating, we find: \[ V = \frac{1}{12}. \] Thus, \( 64V = \frac{64}{12} = \frac{16}{3} \approx 5.33 \). Final Answer: \[ \boxed{5.33}. \]