Question

Show that the surface area of the cone \(z=k \sqrt{x^{2}+y^{2}}, k>0\) over the circular region \(x^{2}+y^{2} \leq r^{2}\) in the \(x y\) -plane is \(\pi r^{2} \sqrt{k^{2}+1}\).

Short answer

The surface area of the cone \(z=k \sqrt{x^{2}+y^{2}}\) over the circular region \(x^{2}+y^{2} \leq r^{2}\) in the \(x y\) -plane is \(πr^2\sqrt{k^2+1}\)

Step-by-step solution

Conversion to Polar Coordinates

Convert the given equation by using polar coordinates as \(x= r \cos θ\) and \(y= r \sin θ\).

Calculating the Integral

The area of the surface element can be described as \(dS= |rdrdθ⋅n| = √{(dr)^2+(rdθ)^2} = √{(1+k^2)r^2}drdθ\). This represents an element of surface area in the polar coordinates on the cone. Here, n is the normal vector. Integral of this surface element over r from 0 to r and over θ from 0 to \(2π\), will give total surface area.

Solving the Integral

Now solve the integral to get the surface area. ∫0ʳ ∫0\(2\pi\) \(r\sqrt{1+k^2}\) dr dθ = \(2π [\frac{r^2\sqrt{k^2+1}}{2}]_0^r\) =\(πr^2\sqrt{k^2+1}\)

Key concepts

Polar Coordinates

Polar coordinates offer an alternative to the more commonly used Cartesian coordinate system, where points are plotted on a grid based on x and y values. Unlike Cartesian coordinates which use horizontal and vertical distances, polar coordinates describe a point on a plane through a straight-line distance from a fixed point, known as the pole (similar to the origin in Cartesian coordinates), and an angle from a direction (typically the positive x-axis).

In mathematical terms, a point \(P\) with polar coordinates \(r, \theta\) relates to Cartesian coordinates as \(x = r \cos \theta\) and \(y = r \sin \theta\). These transformations are especially useful when dealing with curves and shapes that are naturally circular or radial, such as the cone in our exercise. Converting to polar coordinates simplifies the process of integrating over circular regions, which is essential in calculating areas and volumes of such shapes.

Surface Integrals

Surface integrals extend the idea of multiple integrals to more complex surfaces, allowing us to accumulate quantities across a two-dimensional surface in three-dimensional space. Surface integrals can evaluate a variety of things, but in the context of our exercise, we are interested in the surface area of a geometric shape.

To find the surface area of a shape like a cone, you need to compute an integral over the shape's surface. The cone's curved surface can be sliced into infinitesimally small elements, each with an area of \(dS\). The total surface area is then the sum—or integral—of all these tiny areas. The area of each element in polar coordinates takes the form \(dS= \sqrt{(dr)^2+(rd\theta)^2}\), where \( dr\) and \(rd\theta\) represent the infinitesimal changes in the radial and angular directions, respectively. This small section of surface area is calculated considering the gradient of the function describing the cone's surface, resulting in \(dS= \sqrt{1+k^2} r dr d\theta\), which meshes seamlessly with the radial nature of the cone's geometry.

Cone Volume Calculus

The volume of a cone is a fundamental concept in calculus and geometric applications. Calculus provides tools to calculate not only areas but also volumes of various shapes. For a cone, the volume can be computed using integral calculus by slicing the cone into infinitely many thin disks and summing up the volume of each disk.

However, when it comes to surface area, we must remember that we're not dealing with the internal volume but with the area of the cone's ‘skin’. The integral for volume would be a different equation, but for the surface area of a cone, we focus solely on the lateral or slanted surface, excluding the base area, which significantly alters the integral being solved.

Understanding the distinction between calculations for volume and surface area is pivotal. For the volume of a right circular cone with radius \(r\) and height \(h\), the formula is \(\frac{1}{3}\pi r^2h\), which can also be derived using calculus by integrating along the height of the cone. Yet, in the context of our exercise, we're interested in the external surface, which requires understanding and applying concepts of surface integrals.