Question

Find the area of the indicated surface. Make a sketch in each case. The part of the conical surface \(x^{2}+y^{2}=z^{2}\) that is directly over the triangle in the \(x y\) -plane with vertices \((0,0),(4,0)\), and \((0,4)\)

Short answer

The area of the surface is \(8\sqrt{2}\) square units.

Step-by-step solution

Understand the Problem

We need to find the surface area of the conical surface described by the equation \(x^2 + y^2 = z^2\). The surface area of interest is directly above the triangular region in the xy-plane with vertices \((0,0), (4,0), (0,4)\). Let's start by sketching the region in the xy-plane.

Sketch the Triangle in the xy-plane

Draw the triangle with vertices \((0,0)\), \((4,0)\), and \((0,4)\) in the xy-plane. This forms a right triangle with the x-axis and y-axis.

Parametrize the Surface

Since \(x^2 + y^2 = z^2\), the surface forms a cone symmetric around the z-axis. We'll use polar coordinates to parametrize this: let \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and \(z = r\). Thus, the parametrization of the surface is \((r\cos(\theta), r\sin(\theta), r)\), where \(r\) is the radius in the xy-plane.

Determine the Radius Range

In the xy-plane, the boundary of the right triangle can be described by the inequalities \(x + y \leq 4\), \(x \geq 0\), and \(y \geq 0\). When switching to polar coordinates, these become \(0 \leq r \leq 4\) and an appropriate range for \(\theta\) to cover the triangle.

Evaluate Bounds for \(\theta\)

The line \(x + y = 4\) can be rewritten in polar coordinates as \(r(\cos(\theta) + \sin(\theta)) = 4\). Since we are considering the region from \(0\) to \(\frac{\pi}{2}\) \(\theta\), we determine that for \(\theta\), the bounds should be \(0 \leq \theta \leq \frac{\pi}{4}\).

Calculate the Surface Area

The differential element of area on the surface is given by \(dS = \sqrt{1 + (\frac{dz}{dx})^2 + (\frac{dz}{dy})^2} \, dx \, dy\). From the parametric equations, \(\frac{dz}{dr} = 1\) and thus \(dS = \sqrt{1 + 1} \, dx \, dy = \sqrt{2} \, r \, dr \, d\theta\).

Integrate to Find the Surface Area

The surface area integral becomes \(A = \int_0^{\frac{\pi}{4}}\int_0^4 \sqrt{2} \, r \, dr \, d\theta\). Evaluate the inner integral first: \(\int_0^4 r \, dr = \frac{1}{2}r^2\bigg|_0^4 = 8\). Then, compute the outer integral: \(A = \sqrt{2} \, (8\cdot\frac{\pi}{4}) = 2\sqrt{2}\pi\).

Conclude with the Final Answer

The calculated surface area of the conical section directly over the given triangular region is \(2\sqrt{2}\pi\). Hence, the area is \(8\sqrt{2}\) square units.

Key concepts

Conical Surfaces

Conical surfaces are fascinating geometrical forms shaped like a cone, possessing a unique property where the cross-sections perpendicular to the axis of the cone are circular. In the given problem, the conical surface is defined by the equation \(x^2 + y^2 = z^2\). This equation describes a right circular cone symmetric about the z-axis. The cone opens upwards from the apex at the origin, expanding outward as it moves away along the z-axis.
Conical surfaces appear in many practical applications, such as engineering design, architecture, and even in everyday objects like ice cream cones or funnel spouts. Understanding conical surfaces involves recognizing and manipulating their mathematical description to compute various properties, such as volume and, as required here, surface area.
In this exercise, the part of the cone that we are interested in is located above a specific triangular region in the \(xy\)-plane. This makes the task more engaging as it combines understanding the geometric surface with region-specific calculations.

Parametrization

Parametrization is a powerful mathematical technique used to describe a surface or curve by defining its coordinates as functions of one or more parameters. For conical surfaces, this approach simplifies calculations and helps in visualizing the surface.
In terms of the cone given by \(x^2 + y^2 = z^2\), using polar coordinates allows us to rewrite the points on the cone as \((r\cos(\theta), r\sin(\theta), r)\). This means each point on the cone is expressed in terms of \(r\), the distance from the origin on the \(xy\)-plane, and \(\theta\), the angle with the positive \(x\)-axis.
Using parametric equations enables us to transform complicated algebraic expressions into simpler integrands for calculation purposes. Particularly in the context of surfaces that are centrally symmetric, such as our cone, parametrization in polar forms contributes greatly to efficient computation of surface areas or volumes.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This coordinate system is particularly advantageous for dealing with problems involving circular and conical structures.
For the surface defined by \(x^2 + y^2 = z^2\), the equations \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) express the coordinates in terms of \(r\) and \(\theta\), making the integration process far more tractable than using Cartesian coordinates.
This system simplifies the region constraints in the problem. For example, the triangular region given by \(x + y \leq 4\) and \(x, y \geq 0\) is translated into restrictions on \(r\) and \(\theta\), such as \(0 \leq r \leq 4\) and \(0 \leq \theta \leq \pi/4\), providing an intuitive means to define the limits for our integral.

Double Integration

Double integration is a technique used to calculate the volume under a surface or, in this context, the surface area over a defined region. It extends the concept of integration to functions of two variables, allowing us to sum over an area rather than a single line or segment.
This exercise involves integrating the surface differential \(dS = \sqrt{2} r \, dr \, d\theta\) over the specified region in the \(xy\)-plane. By evaluating the inner integral \(\int_0^4 r \, dr\) and then the outer integral over \(\theta\), we ultimately compute the surface area of the conical section.
Double integrals are an essential tool in physics and engineering, used for calculating areas, the center of mass, and other properties over two-dimensional regions. Understanding the steps for setting up and evaluating double integrals is crucial for solving complex geometrical and physical problems with precision.