Use a double integral to find the volume of the indicated solid. $$ \begin{aligned} z & =9-2 y \ x & =4 \ y & =2\end{aligned} $$

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We first identify the region in the $$xy$$-plane and the top surface of the solid. The solid is bounded above by the plane  
$$
z = 9 - 2y,
$$
and laterally by the planes  
$$
x = 4 \quad \text{and} \quad y = 2.
$$
Assuming the solid’s base lies in the $$xy$$-plane (where $$z = 0$$) and that the other boundary faces are on the coordinate planes (i.e. $$x = 0$$ and $$y = 0$$), the region in the $$xy$$-plane is the rectangle  
$$
0 \le x \le 4 \quad \text{and} \quad 0 \le y \le 2.
$$

The volume under the surface $$z = 9 - 2y$$ above this base is given by the double integral
$$
V = \int_{y=0}^{2} \int_{x=0}^{4} \bigl(9-2y\bigr) \,dx\,dy.
$$

We now evaluate the integral step by step.

1. Integrate with respect to $$x$$:
   $$
   \int_{x=0}^{4} \bigl(9-2y\bigr) \,dx = \bigl(9-2y\bigr) \int_{x=0}^{4} dx = \bigl(9-2y\bigr) \times \Bigl[ x \Bigr]_{0}^{4} = 4\,(9-2y).
   $$

2. Substitute the result and integrate with respect to $$y$$:
   $$
   V = \int_{y=0}^{2} 4\,(9-2y) \,dy = 4 \int_{y=0}^{2} (9-2y) \,dy.
   $$

3. Evaluate the integral with respect to $$y$$:
   $$
   \int (9-2y) \,dy = 9y - y^2.
   $$
   Compute from $$y=0$$ to 2:
   $$
   \Bigl[9y-y^2\Bigr]_{0}^{2} = \Bigl(9(2)- (2)^2\Bigr) - \Bigl(9(0)- (0)^2\Bigr) = (18-4) - 0 = 14.
   $$

4. Multiply by 4 to obtain the volume:
   $$
   V = 4 \times 14 = 56.
   $$

Thus, the volume of the solid is  
$$
\boxed{56}.
$$
The volume of the solid is 56 cubic units.

Use a double integral to find the volume of the indicated solid.

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