We are tasked with evaluating the triple integral:

$\iiint_B \left(1 - x^2 - y^2 - z^2 \right) \, dV,$

where $B$ is the region defined by:

$B = \{(x, y, z) \mid x^2 + y^2 + z^2 \leq 9, \, y \geq 0, \, z \geq 0\}.$

This is the portion of the ball $x^2 + y^2 + z^2 \leq 9$ in the first octant (where $y \geq 0$ and $z \geq 0$).

Step 1: Switch to spherical coordinates

In spherical coordinates, the transformations are:

$x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi,$ with: $\rho \geq 0, \quad 0 \leq \phi \leq \frac{\pi}{2}, \quad 0 \leq \theta \leq \frac{\pi}{2}.$

The Jacobian for the spherical transformation is:

$dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

The region $B$ corresponds to $0 \leq \rho \leq 3$, $0 \leq \phi \leq \frac{\pi}{2}$, and $0 \leq \theta \leq \frac{\pi}{2}$.

Step 2: Rewrite the integral

In spherical coordinates, the integrand $1 - x^2 - y^2 - z^2$ becomes:

$1 - \rho^2,$

since $x^2 + y^2 + z^2 = \rho^2$ in spherical coordinates.

The integral becomes:

$\iiint_B \left(1 - x^2 - y^2 - z^2 \right) \, dV = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^3 \left(1 - \rho^2 \right) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 3: Separate the variables

The integral separates into three parts:

1. $\int_0^{\pi/2} \sin\phi \, d\phi$,
2. $\int_0^{\pi/2} d\theta$,
3. $\int_0^3 \left(1 - \rho^2 \right) \rho^2 \, d\rho$.

Step 4: Solve each integral

(a) Compute $\int_0^{\pi/2} \sin\phi \, d\phi$:

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

(b) Compute $\int_0^{\pi/2} d\theta$:

$\int_0^{\pi/2} d\theta = \theta \big|_0^{\pi/2} = \frac{\pi}{2}.$

(c) Compute $\int_0^3 \left(1 - \rho^2 \right) \rho^2 \, d\rho$:

Expand the integrand: $\left(1 - \rho^2 \right) \rho^2 = \rho^2 - \rho^4.$

Now integrate term by term: $\int_0^3 \left(\rho^2 - \rho^4 \right) d\rho = \int_0^3 \rho^2 \, d\rho - \int_0^3 \rho^4 \, d\rho.$

For $\int_0^3 \rho^2 \, d\rho$: $\int_0^3 \rho^2 \, d\rho = \left[\frac{\rho^3}{3}\right]_0^3 = \frac{3^3}{3} - 0 = 9.$

For $\int_0^3 \rho^4 \, d\rho$: $\int_0^3 \rho^4 \, d\rho = \left[\frac{\rho^5}{5}\right]_0^3 = \frac{3^5}{5} - 0 = \frac{243}{5}.$

Thus: $\int_0^3 \left(\rho^2 - \rho^4 \right) d\rho = 9 - \frac{243}{5} = \frac{45}{5} - \frac{243}{5} = -\frac{198}{5}.$

Step 5: Combine results

The total integral is:

$\iiint_B \left(1 - x^2 - y^2 - z^2 \right) \, dV = \left(1\right) \cdot \left(\frac{\pi}{2}\right) \cdot \left(-\frac{198}{5}\right).$

Simplify: $\iiint_B \left(1 - x^2 - y^2 - z^2 \right) \, dV = -\frac{198\pi}{10} = -\frac{99\pi}{5}.$

Final Answer:

$\boxed{-\frac{99\pi}{5}}$

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Let me know if you'd like any clarifications or further explanations!

5 Related Questions:

1. How would the limits of integration change if the region $y \geq 0$ were removed?
2. What is the physical interpretation of the integrand $1 - x^2 - y^2 - z^2$?
3. How would this integral change if the ball's radius were $R$ instead of 3?
4. How do we handle such integrals in Cartesian versus spherical coordinates?
5. Can the integral be solved symbolically without converting to spherical coordinates?

One Tip:

Always double-check the bounds and Jacobian when converting to spherical coordinates to avoid errors in volume scaling!