Problem 17 - Calculus: Early Transcendental Functions

Question

Find the flux of \(\mathbf{F}\) over \(\partial Q\) . $$\begin{aligned} &Q \text { is bounded by } z=\sqrt{x^{2}+y^{2}} \text { and } z=\sqrt{2-x^{2}-y^{2}}\\\ &\mathbf{F}=\left\langle x^{2}, z^{2}-x, y^{3}\right\rangle \end{aligned}$$

Step-by-Step Solution

Answer

The result of the flux would be obtained from the evaluation of the triple integral as computed in Step 3.

Step 1: Compute the Divergence of the Vector Field

First, you want to compute the divergence of the vector field \(\mathbf{F}\), which is the scalar function found by taking the dot product of the del operator and the vector field. In this case, the divergence of \(\mathbf{F}\) is given by \(\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^{2}) + \frac{\partial}{\partial y}(y^{3}) + \frac{\partial}{\partial z}(z^{2}-x) = 2x+y^{2}-1\).

Step 2: Set Up the Triple Integral Using Spherical Coordinates

Now, you express the volume integral in terms of spherical coordinates to simplify the computation. Let \(x = \rho sin\phi cos\theta\), \(y = \rho sin\phi sin\theta\), and \(z = \rho cos\phi\). The triple integral is then \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} [2\rho sin\phi cos\theta + (\rho sin\phi sin\theta)^{2} -1]\rho^{2} sin\phi d\rho d\phi d\theta\).

Step 3: Evaluate the Triple Integral

By evaluating the triple integral, you are effectively computing the flux of \(\mathbf{F}\) over \(\partial Q\). This involves direct computation using the formulae of integration.

Step 4: Conclusion

After evaluating the triple integral in Step 3, the final value obtained is the flux of \(\mathbf{F}\) over \(\partial Q\).

Key Concepts

Divergence Theorem

The Divergence Theorem is a powerful tool in vector calculus, connecting the flow of a vector field through a closed surface to the behavior of the vector field inside the surface. Essentially, it provides a link between surface integrals and volume integrals. This theorem states that the flux of a vector field \(\mathbf{F}\) through a closed surface \(\partial Q\) is equal to the integral of the divergence of \(\mathbf{F}\) over the volume \(Q\) enclosed by the surface. Mathematically, it's expressed as: \[ \int_{\partial Q} \mathbf{F} \cdot d\mathbf{S} = \int_{Q} (abla \cdot \mathbf{F}) \, dV \] Here, \(abla \cdot \mathbf{F}\) represents the divergence of the vector field, which is a scalar function giving the rate at which the field spreads out from a point. This theorem is particularly useful because it can simplify complex surface integrals into potentially easier volume integrals. In our exercise, computing the divergence first helped set up the subsequent steps, allowing us to evaluate the flux via a triple integral.

Triple Integral

Triple integrals are used to extend the concept of integration into three dimensions, evaluating functions over a solid region. They effectively sum up values throughout a given volume, providing a cumulative measure. In terms of the Divergence Theorem, once the divergence \(abla \cdot \mathbf{F}\) is computed, a triple integral is set up over the region \(Q\). For our problem, the triple integral is of the form: \[ \int_{Q} (abla \cdot \mathbf{F}) \, dV \] Carrying out this integration involves integrating with respect to each variable, and the key is setting appropriate limits of integration that correspond to the region \(Q\). Triple integrals allow us to consolidate and include all the differential volume elements throughout our region, effectively enabling us to calculate the total divergence within \(Q\), which is equal to the flux across \(\partial Q\) according to the Divergence Theorem.

Spherical Coordinates

In scenarios involving 3-dimensional regions, using spherical coordinates can greatly simplify the integration process. Spherical coordinates transform the standard \((x, y, z)\) into \((\rho, \phi, \theta)\) where:

• \(\rho\) is the radial distance from the origin,
• \(\phi\) is the azimuthal angle from the positive z-axis,
• \(\theta\) is the polar angle in the xy-plane from the positive x-axis.

In spherical coordinates, the volume element \(dV\) becomes \(\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). In the exercise, using these coordinates aligns nicely with the symmetry of the region \(Q\) defined by the two spheres, \(z=\sqrt{x^2+y^2}\) and \(z=\sqrt{2-x^2-y^2}\). Such conversion ensures our limits of integration are simpler, ranging over typical spherical coordinate bounds and making the integral setup more tractable.