We are tasked with evaluating the iterated integral by converting it into cylindrical coordinates:

$\int_{-2}^{0} \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_0^2 xy^2 \, dz \, dy \, dx$

Step 1: Converting to cylindrical coordinates

In cylindrical coordinates, the following transformations apply:

• $x = r\cos\theta$
• $y = r\sin\theta$
• The volume element $dV$ in cylindrical coordinates is $r \, dz \, dr \, d\theta$

The limits for $x$ and $y$ suggest that the region of integration is a semicircle in the $xy$-plane of radius 2, centered at the origin, above the $x$-axis.

• For $x$ from -2 to 0, we are on the left side of the $xy$-plane, which implies that $\theta$ ranges from $\pi/2$ to $\pi$.
• For $y$ from $-\sqrt{4 - x^2}$ to $\sqrt{4 - x^2}$, this traces the upper and lower boundaries of the semicircle.
• The radial distance $r$ goes from 0 to 2.
• The $z$-limits remain from 0 to 2.

Thus, the transformation to cylindrical coordinates changes the integrand $xy^2$ to $(r\cos\theta)(r^2\sin^2\theta) = r^3\cos\theta\sin^2\theta$.

Step 2: Setting up the integral in cylindrical coordinates

The triple integral in cylindrical coordinates is now:

$\int_{\pi/2}^{\pi} \int_0^2 \int_0^2 r^3 \cos\theta \sin^2\theta \, dz \, dr \, d\theta$

Step 3: Evaluating the integral

1. First, evaluate the $z$-integral:

$\int_0^2 r^3 \cos\theta \sin^2\theta \, dz = 2r^3 \cos\theta \sin^2\theta$

2. Now, evaluate the $r$-integral:

$\int_0^2 2r^3 \cos\theta \sin^2\theta \, dr = 2\cos\theta \sin^2\theta \int_0^2 r^3 \, dr$ The integral of $r^3$ is: $\int_0^2 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} = 4$ Thus, the $r$-integral becomes: $8 \cos\theta \sin^2\theta$

3. Finally, evaluate the $\theta$-integral:

$\int_{\pi/2}^{\pi} 8 \cos\theta \sin^2\theta \, d\theta$ This requires a trigonometric identity to simplify: $\sin^2\theta = 1 - \cos^2\theta$ Therefore: $\int_{\pi/2}^{\pi} 8 \cos\theta \sin^2\theta \, d\theta = 8 \int_{\pi/2}^{\pi} \cos\theta (1 - \cos^2\theta) \, d\theta$ Expanding: $8 \left( \int_{\pi/2}^{\pi} \cos\theta \, d\theta - \int_{\pi/2}^{\pi} \cos^3\theta \, d\theta \right)$ The first integral: $\int_{\pi/2}^{\pi} \cos\theta \, d\theta = \sin\theta \Big|_{\pi/2}^{\pi} = 0 - 1 = -1$ For the second integral $\cos^3\theta$, we use a standard trigonometric identity or reduction formula, leading to: $\int_{\pi/2}^{\pi} \cos^3\theta \, d\theta = -\frac{1}{4}$ Thus, the total becomes: $8(-1 + \frac{1}{4}) = 8 \times \left( -\frac{3}{4} \right) = -6$

Final Answer:

The value of the iterated integral is $\boxed{-6}$.

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Would you like more details or have any questions about the process? Here are some related problems you might find helpful:

1. How do you convert triple integrals to spherical coordinates?
2. How would the limits change if the region were a full circle?
3. What is the significance of the Jacobian in cylindrical coordinates?
4. How do you simplify trigonometric integrals in cylindrical coordinates?
5. Can we use symmetry to reduce the complexity of integrals in specific cases?

Tip: Always pay attention to the region's shape when converting integrals to cylindrical or spherical coordinates.