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Let's break down the problem and verify if the provided solution is correct.

### 1. **Given Function and Region:**

The function to integrate is:

\[
f(x, y, z) = \sqrt{x^2 + y^2 + z^2}
\]

The region \( R \) is the solid hemisphere defined by:

\[
R = \{(x, y, z) \mid x^2 + y^2 + z^2 \leq 4, y \geq 0\}
\]

This describes a solid hemisphere of radius 2, centered at the origin, with the constraint that \( y \geq 0 \), so it's the right half of the hemisphere in the \( xy \)-plane.

### 2. **Spherical Coordinates Transformation:**

In spherical coordinates, the following transformations apply:
- \( x = r \sin \theta \cos \phi \)
- \( y = r \sin \theta \sin \phi \)
- \( z = r \cos \theta \)
- \( r = \sqrt{x^2 + y^2 + z^2} \)

For the given hemisphere, the spherical coordinates would have the ranges:
- \( 0 \leq r \leq 2 \) (since the radius is 2),
- \( 0 \leq \theta \leq \pi \) (spanning the full range of \( z \)-direction),
- \( 0 \leq \phi \leq \frac{\pi}{2} \) (since \( y \geq 0 \), we are confined to the first quadrant in the \( xy \)-plane).

### 3. **Integral Setup in Spherical Coordinates:**

In spherical coordinates, the function \( f(x, y, z) = r \), and the volume element \( dV \) becomes:

\[
dV = r^2 \sin \theta \, dr \, d\theta \, d\phi
\]

Thus, the integral of \( f \) over the hemisphere is:

\[
\int_0^{\frac{\pi}{2}} \int_0^\pi \int_0^2 r \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi = \int_0^{\frac{\pi}{2}} \int_0^\pi \int_0^2 r^3 \sin \theta \, dr \, d\theta \, d\phi
\]

### 4. **Evaluate the Integral:**

First, evaluate the innermost integral with respect to \( r \):

\[
\int_0^2 r^3 \, dr = \left[\frac{r^4}{4}\right]_0^2 = \frac{16}{4} = 4
\]

Next, the integral with respect to \( \theta \):

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

Finally, the integral with respect to \( \phi \):

\[
\int_0^{\frac{\pi}{2}} d\phi = \frac{\pi}{2}
\]

Multiplying all these together gives:

\[
4 \times 2 \times \frac{\pi}{2} = 4\pi
\]

### Conclusion:

The integral over the hemisphere is indeed \( 4\pi \), and the setup provided in the problem is correct.

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Do you want further details or have any questions?

Here are some related questions for you:
1. How do you convert between spherical and Cartesian coordinates for any general point?
2. What is the volume of the hemisphere described in the problem?
3. How would the integral change if the hemisphere were shifted in space?
4. How does the surface area of a hemisphere compare to the volume integral result?
5. What is the physical interpretation of this integral if it represents a gravitational or electrostatic potential?

**Tip:** Always double-check the limits of integration when converting to spherical coordinates, as they can change based on the region's geometry.

Integrate the function f(x, y, z) = √(x² + y² + z²) over the solid hemisphere R = {(x, y, z) | x² + y² + z² ≤ 4, y ≥ 0}. In spherical coordinates, the hemisphere is described as {r, θ, ϕ | 0 ≤ r ≤ 2, 0 ≤ θ ≤ π, 0 ≤ ϕ ≤ π/2}. The integral result is stated to be 4π.

Solve the integration of the function f(x, y, z) = √(x² + y² + z²) over a solid hemisphere using spherical coordinates. Detailed explanation of coordinate transformation and triple integration leading to the final result of 4π. Suitable for advanced high school or college-level calculus.

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