Question

In Exercises 17 and 18 , evaluate \(\int_{S} \int \operatorname{curl} \mathbf{F} \cdot \mathbf{N} d S\) where \(S\) is the closed surface of the solid bounded by the graphs of \(x=4\) and \(z=9-y^{2},\) and the coordinate planes. $$ \mathbf{F}(x, y, z)=x y \cos z \mathbf{i}+y z \sin x \mathbf{j}+x y z \mathbf{k} $$

Short answer

The result of the triple integral of the given vector field over the specified region is \(\frac{512\pi}{5}\).

Step-by-step solution

Transformation Conversion of vector field into divergence form

The divergence for the vector field \( F(x,y,z) = x y \cos z \mathbf{i} + y z \sin x \mathbf{j} + x y z \mathbf{k} \) would be done as follows: \( \nabla \cdot \mathbf{F} = \frac{\partial (x y \cos z)}{\partial x} + \frac{\partial (y z \sin x)}{\partial y} + \frac{\partial (x y z)}{\partial z}\). After doing the above operations we get: \(\nabla \cdot \mathbf{F} = y \cos z + z \sin x + x y\).

Defining the region of Integration

The region S is bounded by the graphs of \(x=4\) and \(z=9-y^{2}\), and the coordinate planes. Thus, the limits for integration in cylindrical coordinates will be: \(0 \leq r \leq 2\) , \(0 \leq \theta \leq 2\pi\) , \(0 \leq z \leq 9-r^{2}\).

Evaluating the triple integral

Now, let's evaluate the triple integral using the divergence theorem: \(\iiint_{S} (\nabla \cdot \mathbf{F}) dV \). By substituting the divergence we calculated in step1 and the coordinates we found in step 2 we have: \( \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{9-r^{2}} ( y \cos z + z \sin x + x y) r dz dr d\theta \). After evaluation we have: \(\frac{512\pi}{5}\).