evaluate the integral ∫∫∫v dzdydx  over the sphere x^2 +y^2 +z^2 =a^2

The given problem is to evaluate the triple integral:

\[
\iiint_V dz\, dy\, dx
\]

over the volume of a sphere \( x^2 + y^2 + z^2 = a^2 \).

### Step 1: Convert to spherical coordinates
The equation of the sphere \( x^2 + y^2 + z^2 = a^2 \) suggests that spherical coordinates are the most suitable for this problem. The relations between Cartesian and spherical coordinates are as follows:

\[
x = r \sin \theta \cos \phi, \quad y = r \sin \theta \sin \phi, \quad z = r \cos \theta
\]

where:
- \( r \) is the radial distance,
- \( \theta \) is the polar angle (measured from the positive \(z\)-axis),
- \( \phi \) is the azimuthal angle (measured from the positive \(x\)-axis).

The volume element in spherical coordinates is:

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

### Step 2: Set the limits of integration
For a full sphere, the limits of integration are:
- \( r \) varies from \( 0 \) to \( a \),
- \( \theta \) varies from \( 0 \) to \( \pi \),
- \( \phi \) varies from \( 0 \) to \( 2\pi \).

### Step 3: Write the integral in spherical coordinates
The triple integral over the volume of the sphere in spherical coordinates becomes:

\[
\iiint_V dz\, dy\, dx = \int_0^{2\pi} \int_0^\pi \int_0^a r^2 \sin \theta \, dr \, d\theta \, d\phi
\]

### Step 4: Evaluate the integral

#### 1. Integrate with respect to \(r\):
\[
\int_0^a r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^a = \frac{a^3}{3}
\]

#### 2. Integrate with respect to \(\theta\):
\[
\int_0^\pi \sin \theta \, d\theta = \left[ -\cos \theta \right]_0^\pi = -(-1 - 1) = 2
\]

#### 3. Integrate with respect to \(\phi\):
\[
\int_0^{2\pi} d\phi = 2\pi
\]

### Step 5: Combine the results
Now, multiply the results of each step:

\[
\frac{a^3}{3} \times 2 \times 2\pi = \frac{4\pi a^3}{3}
\]

### Final Answer:
The value of the triple integral over the volume of the sphere is:

\[
\iiint_V dz\, dy\, dx = \frac{4\pi a^3}{3}
\]

This is the volume of the sphere, which makes sense since the integrand is simply 1 (as we were integrating \( dz \, dy \, dx \), which is the volume element in Cartesian coordinates).

---

Do you want more details or have any further questions?

Here are 5 related questions:
1. How would you evaluate the same integral over a hemisphere?
2. What happens if the limits on \( \theta \) are different, e.g., from \( 0 \) to \( \frac{\pi}{2} \)?
3. How would the integral change if the integrand was \( z \) instead of 1?
4. Can you explain how to derive the volume element \( r^2 \sin \theta \, dr \, d\theta \, d\phi \) in spherical coordinates?
5. How would you calculate the surface area of the sphere using spherical coordinates?

**Tip:** When solving integrals over regions with symmetry, such as spheres, converting to spherical or cylindrical coordinates simplifies the process significantly.

Evaluate the integral ∫∫∫v dzdydx over the sphere x^2 + y^2 + z^2 = a^2

Learn how to evaluate the triple integral ∫∫∫v dzdydx over the volume of a sphere x^2 + y^2 + z^2 = a^2 using spherical coordinates. This step-by-step solution demonstrates key concepts in multivariable calculus, including the conversion to spherical coordinates and integration limits.

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