Understanding the Surface Integral and Divergence Theorem

The problem asks us to calculate the value of the surface integral $\int_S \vec{F} \cdot \vec{N} \, ds$. This integral represents the flux of the vector field $\vec{F}$ passing through the closed surface S. The Divergence Theorem (also known as Gauss's Theorem) offers a direct way to compute this flux by converting it into a volume integral over the region V enclosed by the surface S:

$ \int_S \vec{F} \cdot \vec{N} \, ds = \iiint_V (\nabla \cdot \vec{F}) \, dV $

We will apply this theorem by first finding the divergence of $\vec{F}$, defining the region of integration V, and then evaluating the volume integral.

Calculating the Divergence of the Vector Field

The given vector field is $\vec{F} = 2x^2y\hat{i} - y^2\hat{j} + 4xz^2\hat{k}$. The divergence of a vector field $\vec{F} = P\hat{i} + Q\hat{j} + R\hat{k}$ is calculated as:

$ \nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $

Here, $P = 2x^2y$, $Q = -y^2$, and $R = 4xz^2$. Let's compute the partial derivatives:

• $ \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(2x^2y) = 4xy $
• $ \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(-y^2) = -2y $
• $ \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(4xz^2) = 8xz $

Adding these components gives the divergence:

$ \nabla \cdot \vec{F} = 4xy - 2y + 8xz $

Defining the Region of Integration

The surface S bounds a region V. This region V is located in the first octant ($x \ge 0, y \ge 0, z \ge 0$) and is bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0$, $x = 2$, $y = 0$, $z = 0$. The limits for the volume integral are determined as follows:

• The x-coordinate varies from $0$ to $2$: $0 \le x \le 2$.
• The region projected onto the yz-plane is restricted to the first quadrant ($y \ge 0, z \ge 0$) and lies within the circle $y^2 + z^2 = 9$. This area is a quarter-disk of radius 3. Let's call this region $D_{yz}$.

So, the region V is defined by $0 \le x \le 2$ and $(y, z) \in D_{yz}$, where $D_{yz} = \{ (y, z) \mid y \ge 0, z \ge 0, y^2 + z^2 \le 9 \}$.

Setting Up and Evaluating the Volume Integral

We now need to compute the volume integral of the divergence over the region V:

$ \iiint_V (4xy - 2y + 8xz) \, dV $

This integral can be set up as an iterated integral:

$ \int_0^2 \left( \iint_{D_{yz}} (4xy - 2y + 8xz) \, dy dz \right) \, dx $

To evaluate the double integral over $D_{yz}$, it is helpful to switch to polar coordinates in the yz-plane. We use the transformations $y = r \cos\theta$ and $z = r \sin\theta$. The Jacobian for this transformation is $r$. The region $D_{yz}$ in polar coordinates is described by $0 \le r \le 3$ and $0 \le \theta \le \frac{\pi}{2}$.

Substituting these into the double integral:

$ \iint_{D_{yz}} (4x(r \cos\theta) - 2(r \cos\theta) + 8x(r \sin\theta)) \, (r \, dr \, d\theta) $

$ = \int_0^{\pi/2} \int_0^3 (4xr^2 \cos\theta - 2r^2 \cos\theta + 8xr^2 \sin\theta) \, dr \, d\theta $

Let's evaluate this integral by considering each term separately:

1. Integral of $4xr^2 \cos\theta$:

$ \int_0^{\pi/2} \int_0^3 4xr^2 \cos\theta \, dr \, d\theta = 4x \left( \int_0^3 r^2 \, dr \right) \left( \int_0^{\pi/2} \cos\theta \, d\theta \right) $

Calculate the radial part: $ \int_0^3 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^3 = \frac{3^3}{3} = 9 $.

Calculate the angular part: $ \int_0^{\pi/2} \cos\theta \, d\theta = [\sin\theta]_0^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1 $.

So, this term evaluates to $ 4x \times 9 \times 1 = 36x $.

2. Integral of $-2r^2 \cos\theta$:

$ \int_0^{\pi/2} \int_0^3 -2r^2 \cos\theta \, dr \, d\theta = -2 \left( \int_0^3 r^2 \, dr \right) \left( \int_0^{\pi/2} \cos\theta \, d\theta \right) $

Using the results from above: $ -2 \times 9 \times 1 = -18 $.

3. Integral of $8xr^2 \sin\theta$:

$ \int_0^{\pi/2} \int_0^3 8xr^2 \sin\theta \, dr \, d\theta = 8x \left( \int_0^3 r^2 \, dr \right) \left( \int_0^{\pi/2} \sin\theta \, d\theta \right) $

The radial part is again $ 9 $.

Calculate the angular part: $ \int_0^{\pi/2} \sin\theta \, d\theta = [-\cos\theta]_0^{\pi/2} = -\cos(\frac{\pi}{2}) - (-\cos(0)) = 0 - (-1) = 1 $.

So, this term evaluates to $ 8x \times 9 \times 1 = 72x $.

Combining these results, the double integral over $D_{yz}$ is:

$ \iint_{D_{yz}} (4xy - 2y + 8xz) \, dy dz = 36x - 18 + 72x = 108x - 18 $

Finally, we integrate this expression with respect to x from $0$ to $2$:

$ \int_0^2 (108x - 18) \, dx $

Evaluate the integral:

$ = \left[ 108 \frac{x^2}{2} - 18x \right]_0^2 $

$ = \left[ 54x^2 - 18x \right]_0^2 $

$ = (54(2)^2 - 18(2)) - (54(0)^2 - 18(0)) $

$ = (54 \times 4 - 36) - 0 $

$ = (216 - 36) = 180 $

Conclusion

The calculated value of the surface integral $\int_S \vec{F} \cdot \vec{N} \, ds$ is $180$. This corresponds to the third option provided.