To evaluate the given integral $\iiint xyz \, dx \, dy \, dz$ in the first octant using spherical coordinates, we follow these steps:

Step 1: Understanding the bounds

The integral is bounded by:

1. The planes $x = 0$, $y = 0$, and $z = 0$, which define the first octant.
2. The sphere $x^2 + y^2 + z^2 = 1$, which is a sphere of radius 1 centered at the origin.

In spherical coordinates:

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

Step 2: Define the integration limits

• $\rho$: From 0 to 1 (radius of the sphere).
• $\phi$: From 0 to $\frac{\pi}{2}$ (first octant).
• $\theta$: From 0 to $\frac{\pi}{2}$ (first octant).

Step 3: Transform the integral

Substitute the spherical coordinates into the integral: $\iiint xyz \, dx \, dy \, dz = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 (\rho \sin\phi \cos\theta)(\rho \sin\phi \sin\theta)(\rho \cos\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$

Simplify the integrand: $xyz = \rho^3 \sin^2\phi \cos\phi \cos\theta \sin\theta$ Volume element: $dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ Thus, the integrand becomes: $\rho^5 \sin^3\phi \cos\phi \cos\theta \sin\theta$

Step 4: Separate the integral

The integral can be separated into three parts: $\int_0^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta \cdot \int_0^{\frac{\pi}{2}} \sin^3\phi \cos\phi \, d\phi \cdot \int_0^1 \rho^5 \, d\rho$

Part 1: $\int_0^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta$

Use the substitution $u = \sin\theta$, $du = \cos\theta \, d\theta$: $\int_0^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta = \int_0^1 u \, du = \left[\frac{u^2}{2}\right]_0^1 = \frac{1}{2}$

Part 2: $\int_0^{\frac{\pi}{2}} \sin^3\phi \cos\phi \, d\phi$

Use the substitution $u = \sin\phi$, $du = \cos\phi \, d\phi$: $\int_0^{\frac{\pi}{2}} \sin^3\phi \cos\phi \, d\phi = \int_0^1 u^3 \, du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$

Part 3: $\int_0^1 \rho^5 \, d\rho$

$\int_0^1 \rho^5 \, d\rho = \left[\frac{\rho^6}{6}\right]_0^1 = \frac{1}{6}$

Step 5: Combine the results

Multiply the results of the three integrals: $\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{6} = \frac{1}{48}$

Final Answer:

$\iiint xyz \, dx \, dy \, dz = \frac{1}{48}$

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Related Questions:

1. What is the general formula for volume integrals in spherical coordinates?
2. How does symmetry simplify integrals in spherical coordinates?
3. What are the limits of integration for a full sphere in spherical coordinates?
4. How do you convert between Cartesian and spherical coordinates for a given point?
5. Why does the volume element in spherical coordinates include $\rho^2 \sin\phi$?

Tip:

Always visualize the region of integration to ensure the bounds are correctly defined!