Question

In Exercises 23-28, find the volume of the solid generated by revolving the region about the y-axis. the region enclosed by \(x=\sqrt{5} y^{2}, x=0, y=-1, y=1\)

Short answer

The volume of the solid generated is \(\frac{4 \pi \sqrt{5}}{3} units^{3}\).

Step-by-step solution

Formulate Integral for Volume

The volume V of a solid revolution around the y-axis is calculated using the method of cylindrical shells, as defined by the integral \(V = 2 \pi \int_{a}^{b} x \cdot y' dy\). Here x is the radial distance to the y-axis (given by the function \(x=\sqrt{5} y^{2}\)), y' is the height of the shell (which in this case would be dy, as we are working in terms of y), a and b are the limits of integration, which are the y intervals.

Substitute Given Function into Integral

Substitute \(x= \sqrt{5} y^{2}\) into the volume integral: \(V = 2 \pi \int_{-1}^{1} \sqrt{5} y^{2} dy\).

Simplify and Compute the Integral

The integral becomes \(V = 2 \pi \sqrt{5} \int_{-1}^{1} y^{2} dy\). This can be calculated using power rule, which gives \(V = 2 \pi \sqrt{5} [\frac{y^{3}}{3}]_{-1}^{1} = 2 \pi \sqrt{5} (\frac{1}{3} - \frac{-1}{3})\).

Simplify the Final Answer

Simplifying the equation gives the final volume \(V = \frac{4 \pi \sqrt{5}}{3} units^3\).

Key concepts

Cylindrical Shells Method

The cylindrical shells method is a technique for finding the volume of a solid of revolution when the solid is generated by rotating a region around an axis. Unlike the disk or washer method, which integrates perpendicular to the axis of rotation, the cylindrical shells method integrates parallel to the axis. Imagine wrapping the region around the axis like a label around a can.

This method is particularly useful when the shape is revolving around the y-axis, and it's defined as \(V = 2 \pi \int_{a}^{b} x \cdot y' dy\), where \(x\) represents the radius of the shell at any point \(y\), \(y'\) is the height of the cylindrical shell, and \(a\) and \(b\) are the boundaries of integration. In this specific exercise, \(x=\sqrt{5} y^{2}\) describes the curved boundary of the region, and the rotation around the y-axis produces cylindrical shells as the figure is revolved.

Definite Integral

A definite integral represents a powerful concept in calculus which helps calculate the area under the curve of a function, the total accumulation of a quantity, or in this case, the volume of a solid of revolution. It is expressed as \(\int_{a}^{b} f(x) dx\), with \(a\) and \(b\) being the lower and upper limits of the integration.

In the context of our problem, the integral \(\int_{-1}^{1} \sqrt{5} y^{2} dy\) calculates the volume of each infinitesimally thin cylindrical shell from \(y = -1\) to \(y = 1\), and when we sum all these infinitesimal shells together, we get the total volume of the solid of revolution.

Volume Integral Calculus

Volume integral calculus is a segment of integral calculus used to calculate three-dimensional volumes. To apply this concept to find the volume of a solid of revolution, one must set up an integral that accounts for every infinitesimal volume element that makes up the solid.

In this exercise, the cylindrical shells method frames our volume integral. Each cylindrical shell's volume is accounted for by \(2 \pi \cdot x \cdot y' dy\), and integrating this from \(y = -1\) to \(y = 1\) provides the accumulated volume of the entire solid. This integral calculus approach transforms a complex three-dimensional problem into a manageable one-dimensional integral.

Power Rule Integration

The power rule for integration is a fundamental tool used to compute the antiderivative of a function in the form \(f(y) = y^n\), where \(n\) is a real number. The rule states that \(\int y^n dy = \frac{y^{n+1}}{n+1} + C\), with \(C\) representing the constant of integration.

In our example, we're integrating \(y^2\), so applying the power rule gives us \(\int y^2 dy = \frac{y^{3}}{3} + C\). When computing a definite integral, we evaluate this antiderivative at the upper and lower limits of integration and subtract to find the precise volume of the solid. For our cylindrical shell volume \(V = 2 \pi \sqrt{5} \int_{-1}^{1} y^{2} dy\), the power rule simplifies the process considerably, leading to a straightforward calculation of the volume.