Problem 21 - Single Variable Calculus

Question

In Exercises \(13-34\), find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis. $$ x=\sqrt{4-y^{2}}, \quad x=0, \quad y=0 ; \quad \text { the } y \text { -axis } $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the curves \((x=\sqrt{4-y^2}, x=0, y=0)\) about the \(y\)-axis is \(16\pi\).

Step 1: Set up the boundaries and integral

To apply the disk method to find the volume of the solid of revolution, we need to first determine the boundaries along the \(y\)-axis where the region is bounded. Look at the equation \(x=\sqrt{4-y^2}\), where it intersects \(x=0\) and \(y=0\). It intersects the y-axis \((x=0)\) at \(y=\pm2\) and intersects the x-axis \((y=0)\) at \(x=2\). Now, let's set up the integral for the volume of the solid, using the disk method formula: $$ V = \pi \int_{a}^{b} (f(y))^2 dy $$ where \(V\) is the volume, \(a\) and \(b\) are the boundaries, and \(f(y)\) is the equation describing the region as a function of \(y\). For our region, the boundaries are \(a = -2\) and \(b = 2\), and the function is \(f(y) = \sqrt{4-y^2}\).

Step 2: Integrate

Now that we have the integral set up, we can find the volume by integrating: $$ V = \pi \int_{-2}^{2} (\sqrt{4-y^2})^2 dy $$ Simplify the integrand: $$ V = \pi \int_{-2}^{2} (4-y^2) dy $$ Integrate the polynomial in the integrand: $$ V = \pi \left[ 4y - \frac{1}{3}y^3 \right]_{-2}^{2} $$ Evaluate the integral using the fundamental theorem of calculus: $$ V = \pi \left[(4(2) - \frac{1}{3}(2)^3) - (4(-2) - \frac{1}{3}(-2)^3) \right] $$ Simplify: $$ V = \pi(16 - 0) $$

Step 3: Calculate the volume

Finally, calculate the volume of the solid of revolution: $$ V = 16\pi $$ Thus, the volume of the solid generated by revolving the region bounded by the curves \((x=\sqrt{4-y^2}, x=0, y=0)\) about the \(y\)-axis is \(16\pi\).

Key Concepts

Disk Method

The disk method is an integral calculus technique used to calculate the volume of a solid of revolution. It works by slicing the solid into thin disks perpendicular to the axis of rotation. Each disk has a radius that corresponds to the value of the function being rotated at a specific point. The volume of each disk can be thought of as the volume of a cylinder with a very small height, namely the width of the slice.

To find the total volume of the solid, we add up (integrate) the volumes of these infinitesimally thin disks across the range of our function. The formula for the volume of a single disk is \( \pi r^2h \), where \( r \) is the radius (the function value) and \( h \) is the height (the thickness of the disk), which approaches zero. In our exercise, we used this method to calculate the volume of the solid of revolution by revolving a given region around the y-axis.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional shape around an axis. The shape of the solid is completely determined by the curve being rotated (such as a line, parabola, or other functions) and the axis of rotation. In the provided exercise, the region bounded by \( x=\sqrt{4-y^2} \) and the coordinate axes is revolved around the y-axis, creating a symmetrical solid.

These solids are commonly found in everyday objects like bowls, vases, and bottles. The ability to calculate the volumes of such objects is crucial for many fields including engineering and manufacturing, where material costs and storage capacities are determined based on these calculations.

Integral Calculus

Integral calculus is a branch of mathematics that is concerned with the accumulation of quantities and the area under and between curves. It is one of the two principal operations in calculus, with the other being differential calculus. When we calculate an integral, we are often finding the total accumulation of a function over an interval.

An integral assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. Calculation of volumes using integral calculus involves using the method of slicing (like the disk or washer methods) and adding up the contributions from each slice to get the total volume.

Definite Integral

The definite integral is a key concept in integral calculus. It provides a way to calculate accumulation — whether that's areas under curves or other quantities like volumes — bounded between limits. These limits, often represented by \( a \) and \( b \) as the lower and upper limits of integration, respectively, are the points between which we are integrating over our function.

In our exercise, the integral was taken from \( a = -2 \) to \( b = 2 \) with respect to \( y \) to find the volume of the solid of revolution about the y-axis. The 'definite' aspect of the integral means that we are considering a definite range — we know exactly where to start and stop our calculation, which leads to a specific numerical value as opposed to an indefinite integral, which would result in a function that includes an arbitrary constant.