91Ó°ÊÓ

We're going to set up a triple integral to find the volume of the solid. Let V be the volume of the wedge, which can be represented as: $$V = \iiint_S dV$$ The order of integration we will choose is \(dz\,dx\,dy\), with the region S.

To determine the limits of integration, we'll visualize the three dimensions as a bounding box in our integration and express each variable with the smallest and largest values they can take. 1. For the \(z\)-axis, the minimum value occurs along the plane \(z=0\). The maximum value of z occurs where the equation of the cylinder \(x^2 + y^2 = 4\) is intersecting the plane \(y = -z\). So, let's find the intersection by substituting \(y\) with \(-z\) in the equation of the cylinder: $$x^2 + (-z)^2 = 4 \Rightarrow x^2 + z^2 = 4$$ We will convert these cartesian coordinates into cylindrical coordinates. The radius \(r\) and angle \(\theta\) are determined as follows: $$r = \sqrt{x^2 + y^2} = \sqrt{4} = 2$$ $$\theta = \tan^{-1}\frac{y}{x} \Rightarrow -\pi \leq \theta \leq \pi$$ So our z-limits are \(0 \leq z \leq 2\). 2. For the \(x\)-axis, we notice that x can have any value between the cylinder and the yz-plane, so our \(x\) limits will be \(0 \leq x \leq 2\). 3. Finally, for the \(y\)-axis, we know that the minimum value occurs at \(y = -2\) (from the left-hand side of the cylinder), and the maximum value occurs at \(y = 2\). Therefore, \(y\) limits are \(-2 \leq y \leq 2\).

The triple integral, using the limits of integration we found, is given by: $$V = \int_{-2}^2 \int_{0}^2 \int_{0}^2 dz\,dx\,dy$$

Now, let's evaluate the triple integral: $$V = \int_{-2}^2 \int_{0}^2 \left[ z \right]_0^2 dx\,dy = \int_{-2}^2 \int_{0}^2 2 dx\,dy$$ $$V = \int_{-2}^2 \left[ 2x \right]_0^2 dy = \int_{-2}^2 4 dy$$ $$V = \left[ 4y \right]_{-2}^2 = 4(2) - 4(-2) = 16$$ The volume of the wedge formed is 16 cubic units.