let v=<x^3cosz,1-3x^2ycosz-3yz^2sinx,z^3sinx> be the velocity field. compute the flux of the v across the surface x^2+y+x^2=25 where y>0 and the surface is oriented away from the origin
HINT: Use the Divergence Theorem!

Given the hint to use the Divergence Theorem, let's apply it to compute the flux of the vector field \(\mathbf{v} = \langle x^3 \cos z, 1 - 3x^2y \cos z - 3yz^2 \sin x, z^3 \sin x \rangle\) across the surface \(S\) defined by \(x^2 + y + z^2 = 25\) with \(y > 0\), where the surface is oriented away from the origin.

### Step 1: Divergence Theorem
The Divergence Theorem relates the flux of a vector field \(\mathbf{v}\) across a closed surface \(S\) to the volume integral of the divergence of \(\mathbf{v}\) over the region \(V\) enclosed by \(S\):
\[
\iint_S \mathbf{v} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{v}) \, dV
\]
In this case, \(S\) is the surface \(x^2 + y + z^2 = 25\) with \(y > 0\), and \(V\) is the region inside the surface.

### Step 2: Compute the Divergence of \(\mathbf{v}\)
The divergence of the vector field \(\mathbf{v}\) is given by:
\[
\nabla \cdot \mathbf{v} = \frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_3}{\partial z}
\]
where \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\).

Given \(\mathbf{v} = \langle x^3 \cos z, 1 - 3x^2y \cos z - 3yz^2 \sin x, z^3 \sin x \rangle\):
- \(\frac{\partial v_1}{\partial x} = \frac{\partial}{\partial x} (x^3 \cos z) = 3x^2 \cos z\)
- \(\frac{\partial v_2}{\partial y} = \frac{\partial}{\partial y} \left( 1 - 3x^2y \cos z - 3yz^2 \sin x \right) = -3x^2 \cos z - 3z^2 \sin x\)
- \(\frac{\partial v_3}{\partial z} = \frac{\partial}{\partial z} (z^3 \sin x) = 3z^2 \sin x\)

So, the divergence is:
\[
\nabla \cdot \mathbf{v} = 3x^2 \cos z - 3x^2 \cos z - 3z^2 \sin x + 3z^2 \sin x = 0
\]

### Step 3: Apply the Divergence Theorem
Since the divergence of \(\mathbf{v}\) is zero, the volume integral over \(V\) becomes:
\[
\iiint_V (\nabla \cdot \mathbf{v}) \, dV = \iiint_V 0 \, dV = 0
\]

### Step 4: Conclusion
By the Divergence Theorem, the flux of \(\mathbf{v}\) across the surface \(S\) is zero:
\[
\Phi = \iint_S \mathbf{v} \cdot \mathbf{n} \, dS = 0
\]

So, the flux of the vector field \(\mathbf{v}\) across the surface \(x^2 + y + z^2 = 25\) where \(y > 0\) and the surface is oriented away from the origin is **zero**.

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If you would like more details or have any questions, feel free to ask!

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**Relative Questions:**
1. What are the necessary conditions for the Divergence Theorem to apply?
2. How would the result change if the divergence of the vector field was not zero?
3. What is the physical interpretation of a zero flux across a surface?
4. How would you apply the Divergence Theorem if the surface was not closed?
5. What are some examples of vector fields with non-zero divergence?

**Tip:** When the divergence of a vector field is zero, it indicates that the field is solenoidal, meaning it has no net source or sink within the region. This often simplifies flux calculations.

Learn how to compute the flux of a vector field across a given surface using the Divergence Theorem. Detailed solution for the velocity field \( \mathbf{v} = \langle x^3 \cos z, 1 - 3x^2y \cos z - 3yz^2 \sin x, z^3 \sin x \rangle \) across the surface \( x^2 + y + z^2 = 25 \), oriented away from the origin. Suitable for advanced undergraduate students studying vector calculus.

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