Problem 10
Question
Find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) Parabolic cylinder between planes The surface cut from the parabolic cylinder \(y=x^{2}\) by the planes \(z=0, z=3,\) and \(y=2\)
Step-by-Step Solution
Verified Answer
Parametrization can be \(\mathbf{r}(u, v) = (u, u^2, v)\), with \(-\sqrt{2} \leq u \leq \sqrt{2}\) and \(0 \leq v \leq 3\).
1Step 1: Understanding the surface
We are given a parabolic cylinder described by the equation \(y = x^2\). This surface extends infinitely in the \(z\)-direction unless bounded. However, we are given that it is bound between the planes \(z=0\), \(z=3\), and \(y=2\). Thus, the surface is a section of this parabolic cylinder.
2Step 2: Determine the parameter limits
The range of \(z\) is between 0 and 3, as given by the planes \(z=0\) and \(z=3\). For \(y\), since the cylinder is bounded by \(y=2\), and given that \(y = x^2\), we need \(x^2\leq 2\). Therefore, \(-\sqrt{2} \leq x \leq \sqrt{2}\).
3Step 3: Choose parameters and create parameter functions
We can use \(x\) and \(z\) as parameters for the surface. Let the parameter \(u = x\), which ranges from \(-\sqrt{2}\) to \(\sqrt{2}\), and let \(v = z\), which ranges from 0 to 3. Thus, a parametrization of the surface can be given by the vector function \(\mathbf{r}(u, v) = (u, u^2, v)\), where \(u\) and \(v\) are the parameters.
4Step 4: Verify the parametrization
Our parameter function \(\mathbf{r}(u, v) = (u, u^2, v)\) describes a point on the surface for each \((u, v)\). The \(x\)-coordinate is \(u\), the \(y\)-coordinate is \(u^2\) (which satisfies the equation of the parabolic cylinder \(y = x^2\)), and the \(z\)-coordinate is \(v\), which matches the bounds for \(z\). Therefore, this parametrization is consistent with the description of the surface.
Key Concepts
Parabolic CylinderParameter LimitsVector Function
Parabolic Cylinder
A parabolic cylinder is a type of surface in 3D geometry, defined by a specific equation. For our exercise, the equation is given as \(y = x^2\). This forms a curve when viewed in a 2D plane, but extends infinitely along one axis – here, that axis is the \(z\)-axis. Think of it as a parabolic shape that extends infinitely upward and downward along the \(z\)-axis.
You might imagine it like a "valley" that opens up parallel to the \(z\)-axis. In many practical applications, such surfaces are bounded to give them definite shape and form. In our case, this parabola is bounded between three planes: \(z=0\), \(z=3\), and \(y=2\). These planes essentially "slice" the infinite surface, defining a specific section of the parabolic cylinder for study.
You might imagine it like a "valley" that opens up parallel to the \(z\)-axis. In many practical applications, such surfaces are bounded to give them definite shape and form. In our case, this parabola is bounded between three planes: \(z=0\), \(z=3\), and \(y=2\). These planes essentially "slice" the infinite surface, defining a specific section of the parabolic cylinder for study.
Parameter Limits
In order to parametrize a surface, it is crucial to understand the parameter limits. This identifies the range over which the parameters can vary. For the parabolic cylinder discussed, the bounds are defined as follows:
- The plane \(z=0\) indicates the lower limit for the \(z\) parameter, and the plane \(z=3\) sets the upper limit. Thus, \(z\) ranges from 0 to 3.
- The equation \(y = x^2\) is naturally limited by the plane \(y=2\). Solving \(x^2 \leq 2\) gives us \(-\sqrt{2} \leq x \leq \sqrt{2}\). This translates into bounds for our \(x\) parameter, allowing it to vary from \(-\sqrt{2}\) to \(\sqrt{2}\).
These limits are essential because they allow us to confine our parameterization to the relevant section of the surface, which is especially important in real-world applications, where infinite surfaces don't exist.
- The plane \(z=0\) indicates the lower limit for the \(z\) parameter, and the plane \(z=3\) sets the upper limit. Thus, \(z\) ranges from 0 to 3.
- The equation \(y = x^2\) is naturally limited by the plane \(y=2\). Solving \(x^2 \leq 2\) gives us \(-\sqrt{2} \leq x \leq \sqrt{2}\). This translates into bounds for our \(x\) parameter, allowing it to vary from \(-\sqrt{2}\) to \(\sqrt{2}\).
These limits are essential because they allow us to confine our parameterization to the relevant section of the surface, which is especially important in real-world applications, where infinite surfaces don't exist.
Vector Function
A vector function provides a way of representing a surface in three dimensions. In our exercise, this is achieved through the vector function: \( \mathbf{r}(u, v) = (u, u^2, v) \). This function helps us map each point \((u, v)\) to a specific point on the surface.
- \(u\) corresponds to the \(x\)-coordinate in 3D space.
- \(u^2\) naturally arises as the \(y\)-coordinate, consistent with our parabolic cylinder's defining equation \(y = x^2\).
- \(v\) takes the role of the \(z\)-coordinate, and varies independently from 0 to 3.
This vector function is like a roadmap, giving each point \((u, v)\) a precise location on the bounded surface between the planes. It's a crucial tool in geometry, offering a systematic way to redefine and manipulate shapes and surfaces through mathematics and visualization.
- \(u\) corresponds to the \(x\)-coordinate in 3D space.
- \(u^2\) naturally arises as the \(y\)-coordinate, consistent with our parabolic cylinder's defining equation \(y = x^2\).
- \(v\) takes the role of the \(z\)-coordinate, and varies independently from 0 to 3.
This vector function is like a roadmap, giving each point \((u, v)\) a precise location on the bounded surface between the planes. It's a crucial tool in geometry, offering a systematic way to redefine and manipulate shapes and surfaces through mathematics and visualization.
Other exercises in this chapter
Problem 9
Find a potential function \(f\) for the field \(\mathbf{F}.\) \(\mathbf{F}=e^{y+2 z}(\mathbf{i}+x \mathbf{j}+2 x \mathbf{k})\)
View solution Problem 10
Integrate \(G(x, y, z)=y+z\) over the surface of the wedge in the first octant bounded by the coordinate planes and the planes \(x=2\) and \(y+z=1 .\)
View solution Problem 10
Evaluate $$\iint_{S} \nabla \times(y \mathbf{i}) \cdot \mathbf{n} d \sigma,$$ where \(S\) is the hemisphere \(x^{2}+y^{2}+z^{2}=1, z \geq 0.\)
View solution Problem 10
Evaluate \(\int_{C}(x-y+z-2) d s\) where \(C\) is the straight-line segment \(x=t, y=(1-t), z=1,\) from \((0,1,1)\) to \((1,0,1)\)
View solution