To evaluate the given integral using spherical coordinates, follow these steps:

The integral is:

$\int_{-3}^3 \int_{\sqrt{9-x^2}}^{-\sqrt{9-x^2}} \int_{\sqrt{9-x^2-y^2}}^{0} z \sqrt{x^2 + y^2 + z^2} \, dz \, dy \, dx.$

Step 1: Express the coordinates in spherical terms

In spherical coordinates:

• $x = \rho \sin\phi \cos\theta$
• $y = \rho \sin\phi \sin\theta$
• $z = \rho \cos\phi$
• $x^2 + y^2 + z^2 = \rho^2$
• $dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$

The limits for spherical coordinates are:

• $\rho: [0, R]$ (from the origin to the boundary of the sphere)
• $\phi: [0, \pi]$ (from the positive $z$-axis to the negative $z$-axis)
• $\theta: [0, 2\pi]$ (full rotation around the $z$-axis)

Here, the given sphere has radius $3$, so $R = 3$.

Step 2: Rewrite the integral in spherical coordinates

Substitute the given bounds and integrand into spherical coordinates:

• The integrand $z \sqrt{x^2 + y^2 + z^2}$ becomes: $z \sqrt{x^2 + y^2 + z^2} = (\rho \cos\phi) (\rho) = \rho^2 \cos\phi.$
• The differential $dx \, dy \, dz$ becomes $\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$.

Thus, the integral becomes: $\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{3} \rho^2 \cos\phi \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 3: Separate the integral into independent parts

Separate the variables: $\int_{0}^{2\pi} d\theta \cdot \int_{0}^{\pi} \cos\phi \sin\phi \, d\phi \cdot \int_{0}^{3} \rho^4 \, d\rho.$

Evaluate $\int_{0}^{2\pi} d\theta$:

$\int_{0}^{2\pi} d\theta = 2\pi.$

Evaluate $\int_{0}^{\pi} \cos\phi \sin\phi \, d\phi$:

Use the substitution $u = \sin\phi$, $du = \cos\phi \, d\phi$: $\int_{0}^{\pi} \cos\phi \sin\phi \, d\phi = \int_{0}^{1} u \, du = \left[\frac{u^2}{2}\right]_{0}^{1} = \frac{1}{2}.$

Evaluate $\int_{0}^{3} \rho^4 \, d\rho$:

$\int_{0}^{3} \rho^4 \, d\rho = \left[\frac{\rho^5}{5}\right]_{0}^{3} = \frac{3^5}{5} - \frac{0^5}{5} = \frac{243}{5}.$

Step 4: Combine the results

Multiply the results together: $2\pi \cdot \frac{1}{2} \cdot \frac{243}{5} = \frac{2\pi}{2} \cdot \frac{243}{5} = \frac{243\pi}{5}.$

Final Answer:

$\boxed{\frac{243\pi}{5}}$

Would you like a detailed explanation of any step?

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Related Questions:

1. How do we set limits in spherical coordinates for a general sphere?
2. What substitutions are helpful in evaluating trigonometric integrals in spherical coordinates?
3. How does the Jacobian arise in changing coordinates to spherical?
4. Can this integral be evaluated directly in Cartesian coordinates?
5. What physical scenarios use integrals in spherical coordinates?

Tip:

When converting to spherical coordinates, always remember to multiply by the Jacobian ($\rho^2 \sin\phi$)