Problem 9
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 \(z=4-y^{2}\) by the planes \(x=0, x=2,\) and \(z=0\)
Step-by-Step Solution
Verified Answer
The parametrization is \(\vec{r}(u, v) = (u, v, 4 - v^2)\), with \(0 \leq u \leq 2\) and \(-2 \leq v \leq 2\).
1Step 1: Identify the surface and its constraints
We are asked to find a parametrization of the given surface. The surface described is a section of a parabolic cylinder defined by the equation \(z = 4 - y^2\). It is bounded by the planes \(x = 0\), \(x = 2\), and \(z = 0\).
2Step 2: Determine variable ranges
Given the constraints, the variable \(x\) ranges from \(0\) to \(2\). The variable \(z\) is non-negative and satisfies \(0 \leq z \leq 4 - y^2\). For \(z = 0\), we get \(y^2 = 4\), hence \(y\) ranges from \(-2\) to \(2\).
3Step 3: Set up the parametrization
Since the bounds for \(x\) are up to \(2\), and the parabolic shape \(z = 4 - y^2\) gives restrictions for \(y\) and \(z\), we can set up the parametrization by choosing parameters: - Let \(x = u\), where \(0 \leq u \leq 2\), - Let \(y = v\), where \(-2 \leq v \leq 2\), - Then \(z = 4 - v^2\).
4Step 4: Write the parametrization
The parametrization of the surface based on the ranges determined is:\[ \vec{r}(u, v) = \begin{pmatrix} u \ v \ 4 - v^2 \end{pmatrix} \]where \(0 \leq u \leq 2\), and \(-2 \leq v \leq 2\).
Key Concepts
Parabolic CylinderBoundary Planes3D Parametric Equations
Parabolic Cylinder
A parabolic cylinder is a surface generated by sliding a parabola along a line. In this case, the line is parallel to one of the coordinate axes, not rotating or changing shape. The defining equation of a parabolic cylinder is typically of the form \(z = ax^2\) or \(z = ay^2\). Here, we see it as \(z = 4 - y^2\).
This type of surface gets its name because each cross-section parallel to the cylinder axis looks like a parabola. For our equation \(z = 4 - y^2\), every slice perpendicular to the \(x\)-axis and horizontal to the \(y-z\) plane displays a parabolic shape.
The '4' in the equation pushes the entire parabola up by four units along the \(z\)-axis, and the \(-y^2\) means it opens downward. This orientation plays an important role when we consider boundary conditions or constraints.
This type of surface gets its name because each cross-section parallel to the cylinder axis looks like a parabola. For our equation \(z = 4 - y^2\), every slice perpendicular to the \(x\)-axis and horizontal to the \(y-z\) plane displays a parabolic shape.
The '4' in the equation pushes the entire parabola up by four units along the \(z\)-axis, and the \(-y^2\) means it opens downward. This orientation plays an important role when we consider boundary conditions or constraints.
Boundary Planes
Boundary planes are flat surfaces that limit or confine a solid object or surface in certain directions. In this case, we have three boundary planes: two vertical planes and one horizontal plane.
- The plane \(x = 0\) marks the starting edge on the \(x\)-axis.
- The plane \(x = 2\) signifies the end edge on the \(x\)-axis. Together, these planes create a vertical restriction, confining the surface within these two limits.
- The plane \(z = 0\) lays horizontally, marking the lowest point of the surface, ensuring that the surface doesn’t dip below the \(xy\)-plane. It's crucial for keeping \(z\) non-negative.
3D Parametric Equations
Parametrization is a technique to describe surfaces or curves in terms of parameters. By using parameterized equations, we can control the points on a surface with simple variables. For a 3D surface, these equations take the form \(\vec{r}(u,v) = (x(u,v), y(u,v), z(u,v))\).
In our specific problem, we've selected:
This approach makes it easier to calculate anything from surface area to further geometry properties of the structure, as each point can be uniquely defined by the parameters \(u\) and \(v\). Parametric representations simplify many computational problems in both theoretical and applied mathematics.
In our specific problem, we've selected:
- \(x = u\) as a linear parameter controlling the horizontal limit from \(0\) to \(2\).
- \(y = v\) as managing the vertical span from \(-2\) to \(2\).
- Lastly, \(z = 4 - v^2\), reflecting the vertical height constraint imposed by the parabolic nature of the cylinder.
This approach makes it easier to calculate anything from surface area to further geometry properties of the structure, as each point can be uniquely defined by the parameters \(u\) and \(v\). Parametric representations simplify many computational problems in both theoretical and applied mathematics.
Other exercises in this chapter
Problem 8
Find a potential function \(f\) for the field \(\mathbf{F}.\) \(\mathbf{F}=(y+z) \mathbf{i}+(x+z) \mathbf{j}+(x+y) \mathbf{k}\)
View solution Problem 9
Integrate \(G(x, y, z)=x+y+z\) over the surface of the cube cut from the first octant by the planes \(x=a, y=a, z=a\).
View solution Problem 9
Evaluate \(\int_{C}(x+y) d s\) where \(C\) is the straight-line segment \(x=t, y=(1-t), z=0,\) from \((0,1,0)\) to \((1,0,0)\)
View solution Problem 9
Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq h,\) together with its top, \(x^{2}+y^{2} \leq a^{2}, z=h .\) Let \(\mathbf{F}=-y \mathbf{i}+x \mat
View solution