Question

In Exercises 23-28, find the volume of the solid generated by revolving the region about the y-axis. the region enclosed by the triangle with vertices \((0,1),(1,0),\) and \((1,1)\)

Short answer

The volume of the solid is \(\frac{1}{3}\pi\) units cubed

Step-by-step solution

Determine the bounds

The triangle encloses the region between \(y = 1\) and \(y = 0\). So, the bounds for the integration are \(0\) and \(1\).

Set up volume integration

The solid's volume is given by the integral of \(\pi x^{2} dy\) from \(0\) to \(1\). \(x\) is given by \(1 - y\) which represents the x-coordinate of the boundary of the region at any \(y\) height, as it portrays the line from (1,0) to (0,1). So, the volume \(V\) can be expressed as \(V = \int_{0}^{1} \pi (1 - y)^{2} dy.\)

Evaluate the integral

Now, we need to evaluate the integral \(V = \pi \int_{0}^{1} (1 - 2y + y^{2}) dy\). This separates into \(V = \pi\left[ y - y^{2} + \frac{y^{3}}{3} \right]_{0}^{1}\).

Calculate the volume

Substituting the limits of integration, we get \(V = \pi[(1 - 1 + \frac{1}{3}) - (0 - 0 + 0)] = \frac{1}{3}\pi\).

Key concepts

Integration Bounds

Understanding the integration bounds is crucial when dealing with volume calculations involving solids of revolution. The bounds of an integral define the limits within which the integration takes place. For volume calculations, these boundaries typically correspond to the range over which the solid extends.

In the context of the given exercise, the integration bounds are determined by the vertical limits of the triangle, which are from the line where the triangle intersects the y-axis to the topmost vertex of the triangle. Here, these bounds are between y = 0 and y = 1. It is essential to correctly determine these bounds because they dictate the limits of integration, ensuring the integral calculates the correct volume of the solid.

Volume Integration

The technique of volume integration employs integration to calculate the volume of a solid by adding up infinitely many infinitesimally thin disks or washers. For a solid of revolution, this method rotates a region around a specified axis, and the integral computes the volume of the resulting solid.

In the solved problem, we use the disk method to revolve the triangular region around the y-axis. Through this process, each small disk has a radius equal to the x-coordinates of the region's boundary for a given y-value. The integration adds up the volume of these disks along the y-axis from y = 0 to y = 1, effectively constructing the volume of the entire solid.

Evaluating Integrals

Evaluating an integral is about calculating the area under a curve or the accumulation of quantities across a range. When calculating volumes, this process translates to finding a sum of all the tiny pieces that make up the entire solid.

In our example, we integrate the function \( \pi (1 - y)^{2} \), which represents the area of a cross-sectional disk, across the bounds of integration. By expanding the squared term and integrating term by term, we evaluate the definite integral. The integration process changes algebraic expressions into geometric quantities that represent volumes in this case.

Methods of Finding Volume

There are several methods to find the volume of a solid, particularly in calculus. Two standard methods are the Disc Method and the Washer Method. The method chosen depends on the shape of the solid and how it is generated.

In the disc method, used for the given exercise, solids are formed by revolving a region around an axis, and cross-sections perpendicular to the axis are circles. The volume is the sum of all these individual discs' volumes. The washer method, on the other hand, is used when the solid has a 'hole' in the middle, and cross-sections form a 'washer' or 'ring' shape. Each volume calculation technique is a powerful tool that turns abstract mathematical functions into tangible volumes of three-dimensional objects.