Problem 266 - Advanced Calculus

Question

Compute the integral of the vector field \(\mathrm{F}^{\rightarrow}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) \(=\left(\mathrm{y},-\mathrm{x}, \mathrm{z}^{2}\right)\) over the paraboloid \(\mathrm{z}=\mathrm{x}^{2}+\mathrm{y}^{2}\) with \(0 \leq \mathrm{z} \leq 1\)

Step-by-Step Solution

Answer

The integral of the vector field \(\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z}) = \mathrm{y},-\mathrm{x},\mathrm{z}^{2}\) over the paraboloid \(\mathrm{z}=\mathrm{x}^{2}+\mathrm{y}^{2}\) with \(0 \leq \mathrm{z} \leq 1\) is computed by first parametrizing the surface in cylindrical coordinates as \(\mathrm{r}(\theta, s) = \left(s\cos(\theta), s\sin(\theta), s^2\right)\), where \(0\leq\theta\leq2\pi\) and \(0\leq s\leq 1\). The surface element \(d\mathrm{S}\) is then calculated as \(\left (2s^{2}\cos(\theta),2s^{2}\sin(\theta),s\right )\), and \(\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\cdot d\mathrm{S}\) yielded \(-2s^3\sin(\theta)\cos(\theta) + s^3\). Finally, integration over the surface resulted in \(\frac{1}{2}\pi\).

Step 1: Find a parametrization for the surface

Let's parameterize the paraboloid surface. We can use cylindrical coordinates to parameterize it: $$\mathrm{r}(\theta, s) = \left(s\cos(\theta), s\sin(\theta), s^2\right)$$ where \(0\leq\theta\leq2\pi\) and \(0\leq s\leq 1\).

Step 2: Calculate \(d\mathrm{S}\)

Now, we need to calculate \(d\mathrm{S}\) by taking the cross product of the gradients of the components of r with respect to the parameters: $$\frac{\partial\mathrm{r}}{\partial\theta} = \left(-s\sin(\theta), s\cos(\theta), 0\right)\\ \frac{\partial\mathrm{r}}{\partial s} = \left(\cos(\theta), \sin(\theta), 2s\right)$$ Now, we can compute the cross product: $$d\mathrm{S} = \frac{\partial\mathrm{r}}{\partial\theta} \times \frac{\partial\mathrm{r}}{\partial s} = \left (2s^{2}\cos(\theta),2s^{2}\sin(\theta),s\right )$$

Step 3: Calculate \(\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\cdot d\mathrm{S}\)

Now, compute the dot product of the vector field \(\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\) and \(d\mathrm{S}\): $$\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\cdot d\mathrm{S} = \left(y, -x, z^2\right)\cdot\left(2s^{2}\cos(\theta),2s^{2}\sin(\theta),s\right) \\ = 2s^2y\cos(\theta)-2s^2x\sin(\theta)+sz^2$$ We can replace x, y and z with their corresponding cylindrical representations: $$\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\cdot d\mathrm{S}=(2s^3\sin^2(\theta)\cos(\theta)-2s^3\cos^2(\theta)\sin(\theta)) +s^3 \\ =-2s^3\sin(\theta)\cos(\theta) + s^3$$

Step 4: Calculate the integral

Now, we compute the integral over the surface: $$\iint_{S}{\mathrm{F}^{\rightarrow} \cdot d\mathrm{S}} =\int_{0}^{2\pi}\int_{0}^{1}{\left(-2s^3\sin(\theta)\cos(\theta) + s^3\right) ds d\theta}$$ We can split this into two integrals: $$\int_{0}^{2\pi}\int_{0}^{1}{-2s^3\sin(\theta)\cos(\theta)\,ds\,d\theta} + \int_{0}^{2\pi}\int_{0}^{1}{s^3\,ds\,d\theta}$$ Integrating with respect to s first for each term: $$-\int_{0}^{2\pi}{\left(\frac{1}{2}\sin(\theta)\cos(\theta)\right) d\theta} + \int_{0}^{2\pi}{\left(\frac{1}{4}\right)d\theta}$$ Now, integrating with respect to \(\theta\): $$-\left(\frac{1}{4}\sin^2(\theta)\Big|_0^{2\pi}\right) + \left(\frac{1}{4}\theta\Big|_0^{2\pi}\right) = \frac{1}{2}\pi$$ Thus, the integral of the vector field \(\mathrm{F}^{\rightarrow}(\mathrm{x},\mathrm{y}, \mathrm{z})\) over the paraboloid \(\mathrm{z}=\mathrm{x}^{2}+\mathrm{y}^{2}\) with \(0 \leq \mathrm{z} \leq 1\) is \(\frac{1}{2}\pi\).

Key Concepts

Vector Field

Understanding a vector field is crucial in vector calculus. It is essentially a function that assigns a vector to every point in a subset of space. In this exercise, the vector field given is \( \mathbf{F}(x, y, z) = (y, -x, z^2) \). This means at each point \((x, y, z)\), the vector is composed of components \(y\), \(-x\), and \(z^2\).

• The first component, \(y\), assigns the y-coordinate of the point to the x-component of the vector.
• The second component, \(-x\), assigns the negative of the x-coordinate to the y-component of the vector.
• The third component, \(z^2\), means the z-component of the vector is the square of the z-coordinate.

This particular vector field can represent a variety of physical situations such as velocity fields or force fields around given points in 3-dimensional space.

Parametrization

Parametrization involves expressing a surface or curve in terms of one or more parameters. For surfaces like the paraboloid in our problem, it simplifies calculations such as surface integrals.

For a given surface, the idea is to describe it using a parameterization function, usually denoted by \( r(u, v) \), which maps a region in parameter space (such as \( (u, v) \)) to points on the surface.

In this exercise, we parameterize the paraboloid \(z = x^2 + y^2\) using cylindrical coordinates with parameters \(\theta\) and \(s\), where:
\[ r(\theta, s) = (s\cos(\theta), s\sin(\theta), s^2) \]
Here, \(0 \leq \theta \leq 2\pi\) and \(0 \leq s \leq 1\) define the bounds, neatly covering the surface below \(z = 1\).

Cylindrical Coordinates

Cylindrical coordinates are a natural choice for dealing with problems involving symmetry around an axis, like circular and parabolic structures. These coordinates use a combination of radial distance, angle, and height to describe a point in 3D.

• The coordinate \(s\) represents the radial distance from the z-axis.
• \(\theta\) represents the angle with the positive x-axis in the xy-plane.
• \(z = s^2\) represents the height of the paraboloid in terms of the parameter \(s\).

This system helps simplify the integration process by aligning with the paraboloid's symmetry.

Cross Product

The cross product is an essential tool when working with surface integrals. It allows us to find a vector that is perpendicular to a given surface, which is crucial for finding the differential surface area.

In the parameterized surface \( r(\theta, s) = (s\cos(\theta), s\sin(\theta), s^2) \), we find partial derivatives \( \frac{\partial r}{\partial \theta} \) and \( \frac{\partial r}{\partial s} \) to compute the cross product:

• \( \frac{\partial r}{\partial \theta} = (-s\sin(\theta), s\cos(\theta), 0) \)
• \( \frac{\partial r}{\partial s} = (\cos(\theta), \sin(\theta), 2s) \)

Doing the cross product gives \(d\mathbf{S}\):
\[ d\mathbf{S} = (2s^2\cos(\theta), 2s^2\sin(\theta), s) \] This vector represents the orientation and area of the differential surface element used in our surface integral.

Dot Product

The dot product, or scalar product, is crucial when integrating vector fields over surfaces. It essentially measures how much one vector extends in the direction of another and results in a scalar value.

To compute the surface integral of \( \mathbf{F} \) over the surface, the dot product \( \mathbf{F}(\mathbf{x}, \mathbf{y}, \mathbf{z}) \cdot d\mathbf{S} \) is calculated as: \[ \left(y, -x, z^2\right) \cdot (2s^2\cos(\theta), 2s^2\sin(\theta), s) \] Substituting \(x = s\cos(\theta)\), \(y = s\sin(\theta)\), \(z = s^2\), results in:
\[ -2s^3\sin(\theta)\cos(\theta) + s^3 \] These expressions are used to evaluate the integral over the surface.

Paraboloid Surface

A paraboloid surface is a 3-dimensional shape generated by a quadratic equation, resembling a parabola when sliced along its axis. In this exercise, the paraboloid is described by \( z = x^2 + y^2 \), with 0 ≤ z ≤ 1.

• This is essentially a circular parabola extending upwards, contributing grazing curves that bend outward from its center.
• Its shape makes it ideal for cylindrical parametrization due to rotational symmetry around the z-axis.
• It is common in physics and engineering to model dishes or mirrors that focus energy or particles.

Understanding this geometric shape helps in visualizing and solving problems involving surface integrals.