Problem 40 - Essential Calculus

Question

Determine which value best approximates the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(y\) -axis. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.)\(y=\tan x, y=0, x=0, x=\frac{\pi}{4}\) (a) 3.5 (b) \(-\frac{9}{4}\) (c) 8 (d) 10 (e) 1

Step-by-Step Solution

Answer

The best approximate value for the volume of the solid generated by revolving the bounded region about the y-axis is (a) 3.5.

Step 1: Sketch the given equations

Start by sketching the functions \(y=\tan x\), \(y=0\), \(x=0\) and \(x=\frac{\pi}{4}\) on a coordinate plane. The sketch would show that \(y=\tan x\) raises from 0 to \(\infty\) between \(x=0\) and \(x=\frac{\pi}{4}\), \(y=0\) is a straight line on the x-axis, \(x=0\) is a straight line on the y-axis, and \(x=\frac{\pi}{4}\) is a vertical line crossing the y-axis at \(\frac{\pi}{4}\). The bounded region is a triangle.

Step 2: Visualize the solid

Next, imagine the sketched region being revolved around the y-axis. It would generate a cone-shaped solid.

Step 3: Approximate the volume

Visually, it can be seen that when the solid is formed by revolving around the y-axis, it forms a cone shape. As no calculations are performed, the volume could be approximated fairly small, not as dense as a full cylinder (pi) but smaller than half of pie. The closest approximate value from the given options would be option (a) 3.5.

Key Concepts

calculus

Calculus is a field of mathematics focusing on change and motion. It helps us solve problems related to rates of change (differential calculus) and areas under curves (integral calculus). When dealing with the volume of solids of revolution, calculus becomes particularly useful. The key concept here is integrating a function to find the volume.Integrals can help calculate the area under a curve or a volume when a region gets revolved around an axis. This is known as finding the "volume of a solid of revolution." By visualizing and integrating, we can find precise volumes or even approximate them using simpler geometric shapes or visual approximations.In the problem given, calculus allows us to understand how functions like \(y = \tan x\) interact between specified bounds. These interactions form regions that we can revolutionize around an axis to estimate their volumes.

sketching functions

Sketching functions is a fundamental skill in calculus that helps to visualize mathematical expressions quickly. Patterns become more apparent when equations are sketched, aiding in understanding their behavior between intervals or boundaries.Consider the function \(y=\tan x\). This function increases from 0 to \(\infty\) between \(x=0\) and \(x=\frac{\pi}{4}\). When sketching this, you'll plot these changes on a graph to see precisely how the curve behaves. The line \(y=0\) represents the x-axis, while \(x=0\) and \(x=\frac{\pi}{4}\) indicate vertical lines crossing the y-axis as boundaries.Sketching the bounded region formed by these functions reveals a triangle-like shape. This understanding helps when considering volume, as it provides a clearer picture of the shape to be revolved.

revolving around axis

Revolving a shape around an axis is a technique to create three-dimensional solids. This concept transforms a 2D shape into a 3D solid, which can be useful for real-world applications like engineering and physics.In this problem, the region bounded by \(y=\tan x\), \(y=0\), \(x=0\), and \(x=\frac{\pi}{4}\) is revolved around the y-axis. Visualizing this revolution makes it easier to understand the resulting solid, which in this case is cone-shaped. The lines and curves set the boundaries and outline the resulting 3D object through its revolution.When a triangle or similar shape revolves around the y-axis, it forms a unique 3D shape. Through this process, we explore spatial transformations, which lead us to approximate the volumes without extensive calculations.

visual approximation

Visual approximation is an effective method to estimate the volume or area of complex shapes by simplifying them into recognizable forms. When calculations seem unnecessary or cumbersome, visual approximation fits perfectly. In our particular problem, the obtained shape resembles a cone. Recognizing that it is not a full cylinder nor a complex shape through visual inspection, allows us to come to an educated guess about its volume. Examining the options provided, the visually estimated volume of this cone-like solid is closest to 3.5. Although visual approximations lack the precision of calculus-based computations, they serve excellently for an initial assessment. This demonstrates the power of an intuitive approach to tackling problems involving geometry and calculus.