Problem 9 - Thomas Calculus

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(y\) -axis at \(y=0\) and \(y=2 .\) The cross-sections perpendicular to the \(y\) -axis are cir- cular disks with diameters running from the \(y\) -axis to the parabola \(x=\sqrt{5} y^{2}.\)

Step-by-Step Solution

Answer

The volume of the solid is \(8\pi\) cubic units.

Step 1: Understand the Problem

The problem asks for the volume of a solid with circular cross-sections perpendicular to the y-axis. The diameter of each disk runs from the y-axis to the curve defined by the equation \(x = \sqrt{5}y^2\). The boundaries for integration are given as \(y = 0\) and \(y = 2\).

Step 2: Express the Diameter

The diameter of each disk is the distance from the y-axis to the parabola \(x = \sqrt{5}y^2\). So, the diameter at a particular value of \(y\) is simply \(\sqrt{5}y^2\).

Step 3: Calculate the Radius

Since the radius is half of the diameter, the radius \(r(y)\) can be expressed as \(r(y) = \frac{\sqrt{5}y^2}{2}\).

Step 4: Set Up the Volume Formula

The area of a circular disk is \(A = \pi r^2\). So, the area of the cross-section at any \(y\) is \(A(y) = \pi \left(\frac{\sqrt{5}y^2}{2}\right)^2\).

Step 5: Simplify the Cross-Section Area Formula

Substitute the expression for the radius into the area formula: \[ A(y) = \pi \left(\frac{\sqrt{5}y^2}{2}\right)^2 = \pi \cdot \frac{5y^4}{4} = \frac{5\pi y^4}{4} \]

Step 6: Integrate to Find the Volume

The volume of the solid is the integral of the cross-sectional area along the y-axis, \[ V = \int_{0}^{2} A(y) \, dy = \int_{0}^{2} \frac{5\pi y^4}{4} \, dy \]

Step 7: Perform the Integration

Integrate the function: \[ V = \frac{5\pi}{4} \int_{0}^{2} y^4 \, dy \] Calculate the definite integral of \(y^4\): \[ V = \frac{5\pi}{4} \left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{5\pi}{4} \left( \frac{32}{5} - 0 \right) \] \[ V = \frac{5\pi}{4} \times \frac{32}{5} = 8\pi \]

Step 8: Conclude the Solution

The volume of the solid is computed to be \(8\pi\) cubic units.

Key Concepts

Circular Disk Cross-Section

When dealing with solids in calculus, understanding cross-sections is crucial. A circular disk cross-section is a slice of the solid that takes the shape of a circle. In this exercise, each cross-section of the solid perpendicular to the y-axis is a circular disk.
The diameter of these disks is determined by the range from the y-axis to the parabola described by the equation \( x = \sqrt{5}y^2 \). Because the solid is composed of countless such disks stacked from \( y = 0 \) to \( y = 2 \), understanding this diameter is essential for calculating volume. By recognizing that each cross-section is a disk, we identify a platform to apply further calculus to determine volume.

Integral Calculus

Integral calculus is a branch of calculus focused on accumulation and area under a curve. It allows us to find quantities like areas, volumes, and total values based on rates of change. Here, it serves as a tool to sum up an infinite number of infinitesimally thin slices, or disks, to calculate the volume of the solid.
The definite integral is used over the range of \( y = 0 \) to \( y = 2 \). This matches the region through which the disks extend. By integrating the area function of each disk, we find the entire solid's volume. Calculus helps in turning abstract geometrical ideas into concrete numbers.

Volume by Integration

Calculating volume using integration is a powerful method for finding the volume of complex shapes. In this method, you begin by slicing the solid into known shapes—in this case, circular disks—whose properties you understand.
You then express the volume of each disk as a function of its area (\( A = \pi r^2 \)) and the infinitesimal width along the y-axis (\( dy \)). As a formula, it's written as \( V = \int A(y) \, dy \). In this exercise, we simplified by integrating the cross-sectional area over the range of interest. This involves careful step-by-step work, ensuring every element reflects the geometry of the solid.

Parabola Equation

The parabola equation \( x = \sqrt{5} y^2 \) defines the outer edge of the solid in this exercise. The shape it outlines is crucial in defining the bounds for diameter—and hence radius—of each circular disk.
Understanding this function is key. It means each point on the outer boundary of a disk, for a specific \( y \), will lie on this curve. Because it's a parabola opening to the right, the shape impacts how large each disk gets as \( y \) changes.
Recognizing how the parabola limits and shapes the solid informs every subsequent calculation in the volume integration process.