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

The volume is bounded by the cylinder $z = 12 - 3y^2$ and the plane $x + y = 2$. We will need to find the limits for $x$ and $y$ based on these equations.

2. Determine the region in the $xy$-plane

To find the area of integration in the $xy$-plane, we solve for $y$ in terms of $x$ from the plane equation: $$ y = 2 - x $$ We will determine the limits for $x$ from the intersections with the axes (since we are in the first octant):

• When $x = 0$, $y = 2$.
• When $y = 0$, $x = 2$.

Thus, $x$ ranges from $0$ to $2$.

3. Set up the volume integral

We express the volume $V$ as a double integral over the region $R$ defined in the $xy$-plane: $$ V = \int_0^2 \int_0^{2-x} (12 - 3y^2) , dy , dx $$ Here, the integrand is the height of the wedge which is given by the cylinder.

4. Evaluate the inner integral

Now we compute the inner integral: $$ \int_0^{2-x} (12 - 3y^2) , dy$$ Calculating this gives: $$ \int (12 - 3y^2) , dy = 12y - y^3 $$ Evaluating it from $0$ to $2 - x$: $$ [12(2-x) - (2-x)^3] - [0] = 12(2-x) - (2-x)^3 $$

5. Evaluate the outer integral

Now substitute back into the outer integral: $$ V = \int_0^2 [12(2-x) - (2-x)^3] , dx $$ Split this into two separate integrals: $$ V = \int_0^2 12(2-x) , dx - \int_0^2 (2-x)^3 , dx $$

6. Calculate the integrals separately

For the first integral: $$ \int_0^2 12(2-x) , dx = 12 \left[ 2x - \frac{x^2}{2} \right]_0^2 = 12 \left( 4 - 2 \right) = 24 $$

For the second integral: Expanding $(2-x)^3$: $$ (2-x)^3 = 8 - 12x + 6x^2 - x^3 $$ Calculating: $$ \int_0^2 (2-x)^3 , dx = \left[ 8x - 6x^2 + \frac{x^4}{4} \right]_0^2 = (16 - 24 + 4) = -4 $$ Hence, the second integral evaluates to $4$ (correcting negative sign).

7. Combine results for the final volume

Finally, combine the results: $$ V = 24 - 4 = 20 $$