To evaluate the given integral using cylindrical coordinates, let's break it down step by step:

The integral in Cartesian coordinates:

$\int_{-1}^{1} \int_{\sqrt{1-x^2}}^{-\sqrt{1-x^2}} \int_{2-x^2-y^2}^{2-x^2-y^2} \frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} \, dz \, dy \, dx$

Step 1: Convert to cylindrical coordinates

In cylindrical coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $z = z$
• $x^2 + y^2 = r^2$
• $dx \, dy = r \, dr \, d\theta$

The given integrand simplifies as: $\frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} = \frac{r^2}{r^3} = \frac{1}{r}$

Step 2: Determine the limits of integration

• The domain of integration is a sphere of radius 1 projected in the $xy$-plane. This gives $r \in [0, 1]$ and $\theta \in [0, 2\pi]$.
• $z$ is bounded by $z = 2 - r^2$ and $z = 2 - r^2$ (this implies no variation in $z$, so $z$ contributes nothing to the integration).

Step 3: Set up the integral

The integral in cylindrical coordinates becomes: $\int_{0}^{2\pi} \int_{0}^{1} \int_{2-r^2}^{2-r^2} \frac{1}{r} r \, dz \, dr \, d\theta$

Step 4: Simplify the integrand and integrate

1. The $r$ in the numerator and denominator cancel: $\int_{0}^{2\pi} \int_{0}^{1} \int_{2-r^2}^{2-r^2} 1 \, dz \, dr \, d\theta$

2. The $z$-integral evaluates to 0 because the upper and lower bounds are the same: $\int_{2-r^2}^{2-r^2} 1 \, dz = (2 - r^2) - (2 - r^2) = 0$

3. Since the $z$-integral is 0, the entire integral evaluates to: $0$

Final Answer:

$\boxed{0}$

Would you like further clarifications or a breakdown of specific steps?

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Related Questions:

1. What are the general rules for converting a triple integral to cylindrical coordinates?
2. How does the Jacobian come into play when transforming to cylindrical coordinates?
3. What happens if the bounds of integration in the $z$-direction are non-zero?
4. How do cylindrical coordinates simplify integration in symmetrical domains?
5. What is the significance of spherical versus cylindrical coordinates in triple integrals?

Tip:

Always check for symmetry or boundary conditions in integrals—they often simplify the problem or result in zero contributions!