Math Problem Statement

are these asnwers correct?

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.

Solution

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)=x2+y2+z2f(x, y, z) = \sqrt{x^2 + y^2 + z^2}

The region RR is the solid hemisphere defined by:

R={(x,y,z)x2+y2+z24,y0}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 y0y \geq 0, so it's the right half of the hemisphere in the xyxy-plane.

2. Spherical Coordinates Transformation:

In spherical coordinates, the following transformations apply:

  • x=rsinθcosϕx = r \sin \theta \cos \phi
  • y=rsinθsinϕy = r \sin \theta \sin \phi
  • z=rcosθz = r \cos \theta
  • r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}

For the given hemisphere, the spherical coordinates would have the ranges:

  • 0r20 \leq r \leq 2 (since the radius is 2),
  • 0θπ0 \leq \theta \leq \pi (spanning the full range of zz-direction),
  • 0ϕπ20 \leq \phi \leq \frac{\pi}{2} (since y0y \geq 0, we are confined to the first quadrant in the xyxy-plane).

3. Integral Setup in Spherical Coordinates:

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

dV=r2sinθdrdθdϕdV = r^2 \sin \theta \, dr \, d\theta \, d\phi

Thus, the integral of ff over the hemisphere is:

0π20π02rr2sinθdrdθdϕ=0π20π02r3sinθdrdθdϕ\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 rr:

02r3dr=[r44]02=164=4\int_0^2 r^3 \, dr = \left[\frac{r^4}{4}\right]_0^2 = \frac{16}{4} = 4

Next, the integral with respect to θ\theta:

0πsinθdθ=[cosθ]0π=(11)=2\int_0^\pi \sin \theta \, d\theta = \left[-\cos \theta\right]_0^\pi = -(-1 - 1) = 2

Finally, the integral with respect to ϕ\phi:

0π2dϕ=π2\int_0^{\frac{\pi}{2}} d\phi = \frac{\pi}{2}

Multiplying all these together gives:

4×2×π2=4π4 \times 2 \times \frac{\pi}{2} = 4\pi

Conclusion:

The integral over the hemisphere is indeed 4π4\pi, and the setup provided in the problem is correct.


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.

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Math Problem Analysis

Mathematical Concepts

Spherical Coordinates
Triple Integration
Volume of a Hemisphere

Formulas

x = r sin(θ) cos(ϕ)
y = r sin(θ) sin(ϕ)
z = r cos(θ)
f(x, y, z) = r
dV = r² sin(θ) dr dθ dϕ

Theorems

Spherical Coordinate Transformation
Integration in Spherical Coordinates

Suitable Grade Level

Grades 11-12 (Advanced), College Level Calculus