Problem

Use the Divergence Theorem to calculate the flux of $\mathbf{F}$ across the entire surface of $\mathrm{S}$. where $\mathbf{F}(x, y, z)=(x y+2 x z) \mathbf{i}+\left(x^{2}+y^{2}\right) \mathbf{j}+\left(x y-z^{2}\right) \mathbf{k}$ and $\mathbf{S}$ is the surface bounded by the cylinder $x^{2}+y^{2}=4$ and the planes $z=y-2$ and $z=0$

Answer

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Answer

Final Answer: The flux of the vector field across the entire surface of \(\mathbf{S}\) is \(\boxed{0}\)

Steps

Step 1 :Given the vector field \(\mathbf{F}(x, y, z)=(x y+2 x z) \mathbf{i}+\left(x^{2}+y^{2}\right) \mathbf{j}+\left(x y-z^{2}\right) \mathbf{k}\) and the surface \(\mathbf{S}\) bounded by the cylinder \(x^{2}+y^{2}=4\) and the planes \(z=y-2\) and \(z=0\)

Step 2 :We need to calculate the flux of \(\mathbf{F}\) across the entire surface of \(\mathbf{S}\) using the Divergence Theorem

Step 3 :The Divergence Theorem states that the flux of a vector field across a closed surface is equal to the triple integral of the divergence of the field over the volume enclosed by the surface

Step 4 :The divergence of a vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) is given by \(\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\)

Step 5 :The volume enclosed by the surface is a cylinder of radius 2 and height 2 (from z=0 to z=2). We can use cylindrical coordinates to describe this volume, where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The limits of integration for \(r\) are 0 to 2, for \(\theta\) are 0 to \(2\pi\), and for \(z\) are 0 to \(y-2\)

Step 6 :Calculate the divergence of the given vector field, we get \(\nabla \cdot \mathbf{F} = 3y\)

Step 7 :Convert the divergence to cylindrical coordinates, we get \(\nabla \cdot \mathbf{F} = 3r\sin(\theta)\)

Step 8 :Calculate the triple integral of the divergence over the volume enclosed by the surface, we get 0

Step 9 :Since the result of the triple integral is zero, the flux of the vector field across the entire surface of \(\mathbf{S}\) is zero

Step 10 :Final Answer: The flux of the vector field across the entire surface of \(\mathbf{S}\) is \(\boxed{0}\)

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