The exercise asks to calculate the volume $V_1$ of a sphere centered at the origin $(0, 0, 0)$ with radius 2 using spherical coordinates. Let’s solve this step-by-step using spherical coordinates.

Step 1: Spherical Coordinates

In spherical coordinates: $x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi,$ where:

• $\rho$ is the radial distance ($0 \leq \rho \leq R$),
• $\phi$ is the polar angle ($0 \leq \phi \leq \pi$),
• $\theta$ is the azimuthal angle ($0 \leq \theta \leq 2\pi$).

The volume element $dxdydz$ becomes: $dxdydz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 2: Volume of the Sphere

The sphere $D_1$ has radius $R = 2$. Thus, $\rho$ varies from $0$ to $2$, $\phi$ varies from $0$ to $\pi$, and $\theta$ varies from $0$ to $2\pi$. The volume $V_1$ is given by: $V_1 = \iiint_{D_1} dx\,dy\,dz = \int_0^{2\pi} \int_0^\pi \int_0^2 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Step 3: Evaluate the Integral

1. Integral over $\rho$: $\int_0^2 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^2 = \frac{8}{3}.$

2. Integral over $\phi$: $\int_0^\pi \sin\phi \, d\phi = \left[ -\cos\phi \right]_0^\pi = -\cos\pi - (-\cos 0) = 2.$

3. Integral over $\theta$: $\int_0^{2\pi} d\theta = 2\pi.$

Step 4: Combine the Results

Now, multiply the results from all three integrals: $V_1 = \frac{8}{3} \cdot 2 \cdot 2\pi = \frac{32\pi}{3}.$

Final Answer:

The volume of the sphere is: $V_1 = \frac{32\pi}{3}.$

Would you like a detailed breakdown of any part? Let me know!

5 Related Questions:

1. How are cylindrical coordinates used to find the volume of a cylinder?
2. What is the Jacobian determinant in spherical coordinates?
3. How do spherical coordinates simplify the calculation of integrals for spheres?
4. How is the surface area of a sphere derived using spherical coordinates?
5. What is the difference between cylindrical and spherical coordinate systems?

Tip:

When setting up triple integrals in spherical coordinates, always remember to include the Jacobian term $\rho^2 \sin\phi$ for accurate results.