Question

If \(\mathbf{F}=x \mathbf{i}+y \mathbf{j},\) calculate \(\iint \mathbf{F} \cdot \mathbf{n} d \sigma\) over the part of the surface \(z=4-x^{2}-y^{2}\) that is above the \((x, y)\) plane, by applying the divergence theorem to the volume bounded by the surface and the piece that it cuts out of the \((x, y)\) plane. Hint: What is \(\mathbf{F} \cdot \mathbf{n}\) on the \((x, y)\) plane?

Short answer

\(16\pi\)

Step-by-step solution

- State Divergence Theorem

The Divergence Theorem states that for a vector field \( \mathbf{F} \) and a closed surface \( S \) enclosing a volume \( V \), the following holds:\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \ d\sigma = \iiint_{V} (\abla \cdot \mathbf{F}) \ dV \]

- Compute the Divergence of \( \mathbf{F} \)

Given \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} \), compute the divergence: \( \abla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} = 1 + 1 = 2 \)

- Set Up Volume Integral

Using the Divergence Theorem, we need to evaluate: \[ \iiint_{V} (\abla \cdot \mathbf{F}) \ dV = \iiint_{V} 2 \ dV \]

- Define the Volume

The volume \( V \) is bounded by the surface \( z = 4 - x^2 - y^2 \) and the \( (x,y) \) plane. This is a paraboloid over the circular region \( x^2 + y^2 \leq 4 \).

- Convert to Cylindrical Coordinates

Switch to cylindrical coordinates, where \( x = r \cos\theta, \ y = r \sin\theta, \ z = z, \) and the volume integral converts to: \[ \iiint_{V} 2 \ dV = \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4-r^2} 2 \ r \ dz \ dr \ d\theta \]

- Integrate with Respect to \( z \)

Integrate with respect to \( z \): \[ \int_{0}^{4-r^2} 2 \ r \ dz = 2r \left[ z \right]_{0}^{4-r^2} = 2r (4 - r^2) \]

- Integrate with Respect to \( r \)

Next, integrate with respect to \( r \): \[ \ \int_{0}^{2} 2r (4 - r^2) \ dr = 2 \left[ \int_{0}^{2} (4r - r^3) \ dr \right] \] \[ 2 \left[ \ \frac{4r^2}{2} - \frac{r^4}{4} \ \right]_{0}^{2} \] \[ 2 \left[ 8 - \frac{16}{4} \right] = 2 \left[ 8 - 4 \right] = 2 \cdot 4 = 8 \]

- Integrate with Respect to \( \theta \)

Finally, integrate with respect to \( \theta \: \) \[ \int_{0}^{2\pi} 8 \ d\theta = 8\theta \bigg|_{0}^{2\pi} = 8(2\pi - 0) = 16\pi \]

Key concepts

Vector Field

A vector field is a function that assigns a vector to every point in a space. In your exercise, the given vector field is \(\textbf{F} = x \textbf{i} + y \textbf{j}\). This vector field can be visualized as arrows pointing in the direction of \(\textbf{i}\) and \(\textbf{j}\) components across the \((x, y)\) plane. Each vector's tail is placed at its corresponding \((x, y)\) coordinate, and its length and direction vary depending on the values of \x\ and \y\. Understanding vector fields is critical because they are used to model various physical phenomena, such as fluid flow and electromagnetic fields.

Surface Integral

A surface integral is a way to integrate over a surface in three-dimensional space. In this context, the surface integral \(\textbf{F} \cdot\ \textbf{n} \ d \sigma\) represents the flux of the vector field \(\textbf{F}\) across the surface. The surface you’re working with in the exercise is part of the paraboloid defined by \z = 4 - x^2 - y^2\. The surface integral calculates how much of the vector field flows through this surface. The vector \textbf{n}\ represents the normal vector to the surface at each point, and \d \sigma\ represents an infinitesimal piece of the surface. In the Divergence Theorem, surface integrals are equated to volume integrals, simplifying computations.

Cylindrical Coordinates

Cylindrical coordinates are a coordinate system that extends two-dimensional polar coordinates by adding a third dimension, \z\. They describe a point in space by three values: the radial distance \r\ from the \z\-axis, the angular coordinate \theta\ around the \z\-axis, and the height \z\ along the \z\-axis. In the provided problem, switching to cylindrical coordinates simplifies the integration process. In these coordinates, \x = r \cos \theta\, \y = r \sin \theta\, and \z = z\. The volume integral in cylindrical coordinates becomes \(\textbackslash iiint_{V} 2 \ dV = \int_{0}^{2\textbackslash pi} \int_{0}^{2} \int_{0}^{4-r^2} 2 \ r \ dz \ dr \ d \theta\). This conversion simplifies the boundaries and integrand, making the integration more manageable.

Volume Integral

A volume integral is used to find quantities that are accumulated throughout a volume in space. In the context of the Divergence Theorem, the volume integral converts a surface integral over a closed surface into a volume integral over the region bounded by the surface. In the problem, the surface integral \(\textbf{F} \cdot \textbf{n} \ d \sigma\) is transformed to a volume integral using \(\textbackslash iiint_{V}(abla \cdot \textbf{F}) \ dV\). The divergence \(abla \cdot \textbf{F} = 2\) simplifies the volume integral, representing how the vector field diverges within the volume. The bounded volume is between the surface \z = 4 - x^2 - y^2\ and the plane \z = 0\. Integrating in cylindrical coordinates further simplifies the problem, finalizing the integral into a manageable form.