To express and evaluate the given triple integral $\iiint_E 3y \, dV$, where $E$ is the portion of the ball $x^2 + y^2 + z^2 \leq 1$ that lies in the first octant, we can use spherical coordinates.

Step 1: Define the Region $E$ in Spherical Coordinates

In spherical coordinates:

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

The inequality $x^2 + y^2 + z^2 \leq 1$ becomes $\rho \leq 1$, since $\rho^2 = x^2 + y^2 + z^2$.

Since we are considering only the first octant, the angles range as follows:

• $\theta$ from $0$ to $\frac{\pi}{2}$ (for the first quadrant in the $xy$-plane)
• $\phi$ from $0$ to $\frac{\pi}{2}$ (for the positive $z$-axis direction)

Thus, the region $E$ in spherical coordinates is defined by:

• $0 \leq \rho \leq 1$
• $0 \leq \theta \leq \frac{\pi}{2}$
• $0 \leq \phi \leq \frac{\pi}{2}$

Step 2: Rewrite the Integrand $3y$ in Spherical Coordinates

Since $y = \rho \sin \phi \sin \theta$, the integrand $3y$ becomes: $3y = 3 \rho \sin \phi \sin \theta$

Step 3: Set Up the Integral in Spherical Coordinates

Now, we can write the triple integral as: $\iiint_E 3y \, dV = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 3 \rho \sin \phi \sin \theta \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

Simplifying the integrand: $= \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 3 \rho^3 \sin^2 \phi \sin \theta \, d\rho \, d\phi \, d\theta$

Step 4: Evaluate the Integral

Integrate with Respect to $\rho$:

$\int_0^1 3 \rho^3 \, d\rho = 3 \left[ \frac{\rho^4}{4} \right]_0^1 = 3 \cdot \frac{1}{4} = \frac{3}{4}$

Integrate with Respect to $\phi$:

$\int_0^{\frac{\pi}{2}} \sin^2 \phi \, d\phi$ Using the identity $\sin^2 \phi = \frac{1 - \cos(2\phi)}{2}$, we get: $= \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\phi)}{2} \, d\phi = \frac{1}{2} \left[ \phi - \frac{\sin(2\phi)}{2} \right]_0^{\frac{\pi}{2}}$ Evaluating, we have: $= \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$

Integrate with Respect to $\theta$:

$\int_0^{\frac{\pi}{2}} \sin \theta \, d\theta = \left[ -\cos \theta \right]_0^{\frac{\pi}{2}} = -(0 - (-1)) = 1$

Step 5: Combine Results

Multiplying the results of each integral: $\frac{3}{4} \cdot \frac{\pi}{4} \cdot 1 = \frac{3\pi}{16}$

Final Answer

$\iiint_E 3y \, dV = \frac{3\pi}{16}$

Would you like further details on this calculation or have any questions?

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Additional Related Questions

1. How would the limits change if the region were in the entire ball, not restricted to the first octant?
2. Can we evaluate a similar integral if $E$ were bounded by a different shape, like a hemisphere?
3. How can spherical coordinates help simplify integrals over spherical or cylindrical regions?
4. What other transformations can be used for integrals over regions symmetric about the origin?
5. How would the evaluation differ if the integrand were $3x$ instead of $3y$?

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Tip: In spherical coordinates, integrals over symmetric regions often simplify due to the alignment of spherical coordinates with natural symmetries in three-dimensional space.