Find the volume of the wedge cut from the first octant by the cylinder z = 12 - 3y^2 and the plane x + y = 2.

Steps to Solve

1. Identify the boundaries for (x) and (y)

The plane (x + y = 2) limits the region in the (xy)-plane. We can express (x) in terms of (y): $$ x = 2 - y $$ Since we are in the first octant, both (x) and (y) must be non-negative. Thus, (0 \leq y \leq 2).

2. Determine the limits for (z)

The volume is bounded above by the surface defined by the equation of the cylinder: $$ z = 12 - 3y^2 $$ This determines the height of the volume above any point ((x,y)) in the specified region.

3. Set up the integral for the volume

The volume (V) can be calculated using a double integral: $$ V = \int_0^2 \int_0^{2-y} (12 - 3y^2) , dx , dy $$ Here, (x) ranges from (0) to (2 - y) and (y) ranges from (0) to (2).

4. Evaluate the inner integral

First, integrate with respect to (x): $$ \int_0^{2-y} (12 - 3y^2) , dx = (12 - 3y^2) \cdot (2 - y) $$ This expands to: $$ = (12 - 3y^2)(2 - y) = 24 - 12y - 6y^2 + 3y^3 $$

5. Evaluate the outer integral

Now, integrate the result from the previous step with respect to (y): $$ V = \int_0^2 (24 - 12y - 6y^2 + 3y^3) , dy $$

6. Calculate the integral

Calculating each term gives: $$ V = [24y - 6y^2 - 2y^3 + \frac{3}{4}y^4]_0^2 $$ Evaluating this from (0) to (2): $$ = [48 - 24 - 16 + 12] - [0] = 20 $$