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To solve the given integral by changing to spherical coordinates, we need to follow these steps:

1. **Understand the Region**: The integral bounds suggest a volume in 3D space. The integration limits indicate that this is a part of a sphere or semi-sphere.

2. **Spherical Coordinates Conversion**:
   - Spherical coordinates are given by:
     \[
     x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi
     \]
     and the Jacobian determinant for spherical coordinates is:
     \[
     dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.
     \]

3. **Integral Expression in Spherical Coordinates**:
   The integrand \( xy \) transforms as:
   \[
   xy = (\rho \sin\phi \cos\theta)(\rho \sin\phi \sin\theta) = \rho^2 \sin^2\phi \cos\theta \sin\theta.
   \]

4. **Bounds in Spherical Coordinates**:
   - \(\rho\): Radial distance. The limits for \(\rho\) will go from 0 to the radius of the sphere.
   - \(\phi\): Polar angle (measured from the positive \(z\)-axis). For a hemisphere (upper region), \(\phi\) goes from \(0\) to \(\pi/2\).
   - \(\theta\): Azimuthal angle (measured in the \(xy\)-plane). This ranges from \(0\) to \(2\pi\).

   Since the \(x^2 + y^2 + z^2 = \rho^2\), we infer the sphere's radius \(R = 4\).

5. **Set up the Integral**:
   Transform the integral:
   \[
   \int_0^{2\pi} \int_0^{\pi/2} \int_0^4 \rho^2 \sin^2\phi \cos\theta \sin\theta \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.
   \]

   Simplify:
   \[
   \int_0^{2\pi} \int_0^{\pi/2} \int_0^4 \rho^4 \sin^3\phi \cos\theta \sin\theta \, d\rho \, d\phi \, d\theta.
   \]

6. **Separate the Integral**:
   The integral splits into three parts:
   \[
   \int_0^4 \rho^4 \, d\rho, \quad \int_0^{\pi/2} \sin^3\phi \, d\phi, \quad \int_0^{2\pi} \cos\theta \sin\theta \, d\theta.
   \]

7. **Evaluate Each Part**:
   - \(\int_0^4 \rho^4 \, d\rho = \left[\frac{\rho^5}{5}\right]_0^4 = \frac{4^5}{5} = \frac{1024}{5}\).
   - \(\int_0^{\pi/2} \sin^3\phi \, d\phi = \int_0^{\pi/2} \sin\phi (1 - \cos^2\phi) \, d\phi = \left[-\cos\phi + \frac{\cos^3\phi}{3}\right]_0^{\pi/2} = 1 - 0 = 1\).
   - \(\int_0^{2\pi} \cos\theta \sin\theta \, d\theta = \int_0^{2\pi} \frac{1}{2} \sin(2\theta) \, d\theta = 0\), because \(\sin(2\theta)\) is periodic over \([0, 2\pi]\).

8. **Final Result**:
   Since one of the parts evaluates to 0, the entire integral is:
   \[
   0.
   \]

Would you like a deeper explanation of any step or have further questions? Here are some related topics:

1. How to convert integrals to spherical coordinates in more complex regions.
2. Exploring Jacobians in other coordinate systems like cylindrical coordinates.
3. Evaluating trigonometric integrals in spherical coordinates.
4. Using symmetry arguments to simplify integrals.
5. Applications of spherical coordinates in physics and engineering.

**Tip**: Always check for symmetry in the integrand and region—it can often simplify your computations significantly!

Evaluate the integral by changing to spherical coordinates: ∫∫∫ xy dz dy dx over the region √(16−x²) to √(32−x²−y²) and √(x²+y²) with bounds provided.

Learn how to evaluate a triple integral ∫∫∫ xy dz dy dx by converting to spherical coordinates. This problem involves understanding spherical transformations, Jacobians, and simplifying trigonometric integrals. Suitable for undergraduate multivariable calculus students.

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