Question

Order the surfaces in ascending order of the lateral surface area under the surface and over the curve \(y=\sqrt{x}\) from (0,0) to (4,2) in the \(x y\) -plane. Explain your ordering without doing any calculations. (a) \(z_{1}=2+x\) (b) \(z_{2}=5+x\) (c) \(z_{3}=2\) (d) \(z_{4}=10+x+2 y\)

Short answer

The order of surfaces is z3 < z1 < z2 < z4. That is, Surface 3 has the smallest surface area, followed by Surface 1, Surface 2, and Surface 4 has the largest surface area.

Step-by-step solution

Identify Surface Value

Identify the surface value at the extreme points. For all given surfaces, check the value at (0,0) and at (4,2).

Observations

We observe that, at (0, 0), all surfaces have the value as z1=2, z2=5, z3=2 and z4=10. Similarly, at (4,2), calculate the surface value, we get z1=6, z2=9, z3=2 and z4=18.

Ordering

The surface with the least value at (0,0) and (4,2) has the lowest surface area, whereas higher value means more lateral area. So, the order is z3 < z1 < z2 < z4. That is, from smallest to largest surface area, the order is Surface 3, Surface 1, Surface 2, and Surface 4.

Key concepts

Surface Area Integration

Understanding how to integrate a function to find the surface area is an essential skill in calculus. To find the lateral surface area of an object, you need to use surface area integration. This involves setting up an integral that spans the surface of the object. In simple cases, you might use the formula \[ \int_{a}^{b} 2\pi\text{ radius of rotation}(x) \sqrt{1 + (\frac{dy}{dx})^2} dx \] for objects of revolution. More complex surfaces require tailored integrals that account for their specific geometry.

To find the surface area of a lateral side defined over a curve on the plane, and bounded by two specific values of x, you would set up an integral along that curve. If the curve is defined by a function in the form of \( y = f(x) \), and the surface equation is given by \( z = g(x, y) \), you'd need to find an expression for \( dS \), the infinitesimal surface area element, and integrate over the region of interest. This is typical for many calculus surface problems which challenge students to think spatially about curves and surfaces.

Calculus Surface Problems

Calculus problems involving surfaces require a thorough understanding of differentiation and integration. For a student tackling calculus surface problems, visualizing the problem is key. Often, these problems involve working out the area of curved surfaces where you need to consider the rates of change at each point on a surface. This is why derivatives come into play. When you're calculating the lateral surface area, for instance, you often need to integrate a derivative along the axis of the curve.

Understanding Derivatives in Surface Problems

Derivatives represent how a surface changes at a point, and they help in understanding the slope or gradient at any point on the surface. Integrating these derivatives along the curve's boundary helps you find the overall surface area. It's like adding up infinitesimally small 'patches' of area to get the total.

Comparing Surface Areas

Comparing surface areas without actually computing them can often be done by inspecting the defining functions for the surfaces. When given several surface equations, such as \( z = f(x, y) \), you can sometimes determine relative sizes by examining the function's values over the region of interest.

The original exercise showed us that different surfaces can be ordered by their altitudes above the curve \( y = \sqrt{x} \). In practical terms, the higher the surface above the curve, the greater the lateral area between the surface and the curve. This is due to the surface 'stretching' further from the curve to achieve a greater altitude, hence covering more area. By simply comparing the height functions at key points, students can learn to make educated comparisons of surface areas without delving into complex computations.

Curves and Surfaces Calculus

The intersection of curves and surfaces calculus is a fascinating area that combines two-dimensional curves with three-dimensional surfaces to determine areas and volumes. When you're dealing with a curve in the plane and a surface that lies above it, the curve often serves as a boundary for computing the surface area.

Curve as a Boundary

Take, for example, the curve \( y = \sqrt{x} \) from the original problem; it acts as a 'ground level' for the surfaces above. To find the lateral area between this ground and the surface, you'd technically need to project the curve into the third dimension and calculate the area of the 'wall' that forms between the curve and the surface. This involves a deeper understanding of three-dimensional calculus, including partial derivatives and double integrals, where you'd calculate the area over both the x and y coordinates.

For students, visualizing how a curve in the xy-plane constrains a surface in three dimensions is crucial — it's the foundation upon which they'll build more advanced skills in analyzing the geometry of curves and surfaces.