Problem 68 - Calculus Single Variable

Question

In each of Exercises 65-68, use the method of cylindrical shells to calculate the volume obtained by rotating the given planar region \(\mathcal{R}\) about the given line \(\ell\) \(\mathcal{R}\) is the region between the curves \(x=y^{3}\) and \(x=-y^{2} ; \ell\) is the line \(y=3\).

Step-by-Step Solution

Answer

The volume is \(\frac{3\pi}{5}\).

Step 1: Identify the region boundaries

The region \(\mathcal{R}\) is defined by the curves \(x = y^3\) and \(x = -y^2\). We need to find their intersection points by setting \(y^3 = -y^2\). Factor to get \(y^2(y + 1) = 0\), giving us \(y = 0\) and \(y = -1\) as the intersection points. This defines the region between \(y = -1\) and \(y = 0\).

Step 2: Setup the cylindrical shell formula

To find the volume using the method of cylindrical shells, identify the radius and height of a representative shell. The radius is the distance from the shell to the line of rotation \(y = 3\), which is \((3 - y)\). The height is the difference in the x-values of the curves, \(y^3 - (-y^2) = y^3 + y^2\).

Step 3: Write the integral expression

The formula for the volume \(V\) using cylindrical shells is given by: \[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy \]Plugging the radius \((3 - y)\) and the height \(y^3 + y^2\) into the integral from \(y = -1\) to \(y = 0\), we get: \[ V = 2\pi \int_{-1}^{0} (3-y)(y^3 + y^2) \, dy \]

Step 4: Expand the integrand

Expand the expression inside the integral: \((3-y)(y^3 + y^2) = 3y^3 + 3y^2 - y^4 - y^3 = 2y^3 + 3y^2 - y^4\).

Step 5: Integrate term by term

Integrate each term separately over \([-1, 0]\):1. \(\int 2y^3 \,dy = \frac{1}{2}y^4\)2. \(\int 3y^2 \,dy = y^3\)3. \(\int -y^4 \,dy = -\frac{1}{5}y^5\)So the integrated expression becomes: \[\left[\frac{1}{2}y^4 + y^3 - \frac{1}{5}y^5\right]_{-1}^{0}\]

Step 6: Evaluate the integral at limits

Plug in the limits of integration: - At \(y = 0\): \(\frac{1}{2}(0)^4 + (0)^3 - \frac{1}{5}(0)^5 = 0\).- At \(y = -1\): \(\frac{1}{2}(-1)^4 + (-1)^3 - \frac{1}{5}(-1)^5 = \frac{1}{2} - 1 + \frac{1}{5}\).Calculate this as: \(\frac{1}{2} + \frac{1}{5} - 1 = -\frac{3}{10}\).

Step 7: Compute the volume

Subtract the lower limit value from the upper limit:\[ \text{Volume} = 2\pi (0 - (-\frac{3}{10})) = \frac{3\pi}{5} \]This is the volume of the solid obtained by rotating the region \(\mathcal{R}\) about the line \(y = 3\).

Key Concepts

Volume of Solids of Revolution

When studying calculus, a fascinating application is finding the volume of solids of revolution. This involves rotating a two-dimensional area around a line (the axis of rotation) to form a three-dimensional solid. One effective method to find such volumes is the cylindrical shells method. This technique is best used when the solid is generated by rotating around an axis parallel to the axis of the function. Instead of slicing the solid, you "unwrap" it to view as thin, hollow cylinders.

The volume of each cylindrical shell is determined using the formula \[V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy\]- **Radius:** It is the distance from the shell to the axis of rotation.- **Height:** This is typically the difference between functions or "curves" that form the boundaries of the region.

By integrating these along the desired bounds, you find the cumulative volume of all these shells. It provides a practical, calculable approach very useful in scenarios with rotational symmetry.

Integration Techniques

Integration is a core technique in calculus that allows us to calculate areas, volumes, and other quantities that are additive in nature. When dealing with cylindrical shells, the integration process involves evaluating a definite integral that represents the volume of the solid. The basic idea is to break down a complex volume into simpler, manageable pieces and sum these contributions.

First, expand the expression inside the integral, as seen by expanding \((3-y)(y^3 + y^2)\) into \(2y^3 + 3y^2 - y^4\). Integrate each term individually. For instance:

• \(\int 2y^3 \, dy = \frac{1}{2}y^4\)
• \(\int 3y^2 \, dy = y^3\)
• \(\int -y^4 \, dy = -\frac{1}{5}y^5\)

Combining these gives the antiderivative needed to evaluate the volume formula from the lower limit to the upper limit. This process shows how integration methodically helps solve these volumes challenges through structured mathematical processes.

Intersection of Curves

Understanding where curves intersect is crucial when setting up integrals, especially in volume problems. The intersection points define the limits of integration for considerations of area and volume. Curves intersect where their equations equal each other. For the problem at hand, curves \(x = y^3\) and \(x = -y^2\) meet where these equations are equal.

To find these points, solve \(y^3 = -y^2\) by factoring:\(y^2(y + 1) = 0\), which gives solutions \(y = 0\) and \(y = -1\). These points \(-1\) and \(0\) mark the boundaries of integration. Clearly identifying these intersections ensures the integration is bounded by the correct limits, preventing calculation errors and ensuring accuracy in determining the volume formed by rotating the region defined between these intersecting curves.