Problem 23 - Calculus Single Variable

Question

In each of Exercises 19-24, use the method of washers to calculate the volume \(V\) obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\) -axis. \(\mathcal{R}\) is the region in the first quadrant that is bounded on the left by \(y=4 x\), on the right by \(y=x^{2}\), and above by \(y=4\).

Step-by-Step Solution

Answer

The volume \(V\) is \(\frac{20\pi}{3}\).

Step 1: Understand the Boundary Curves

Identify the boundary curves of the region \(\mathcal{R}\). The region \(\mathcal{R}\) is bounded by \(y = 4x\), \(y = x^2\), and \(y = 4\). We are to rotate this region about the \(y\)-axis to find the volume.

Step 2: Find Points of Intersection

To determine the limits of integration, find the points of intersection of the curves \(y = 4x\) and \(y = x^2\). Set \(4x = x^2\). Solving for \(x\), we get \(x(x-4) = 0\), so \(x = 0\) and \(x = 4\). Substituting back to find \(y\)-values, we get intersections at \((0,0)\) and \((2,4)\).

Step 3: Set Up Washer Method Formula

The volume \(V\) for rotating about the \(y\)-axis is calculated using the washer method: \[ V = \pi \int_{y_{min}}^{y_{max}} \left[ (R(y))^2 - (r(y))^2 \right] \ dy \] where \(R(y)\) is the outer radius and \(r(y)\) is the inner radius.

Step 4: Determine Radii in Terms of \(y\)

For \(y\) between 0 and 4, the outer radius \(R(y)\) is derived from \(x\) in \(y = x^2\), giving \(x = \sqrt{y}\). The inner radius \(r(y)\) comes from solving \(y = 4x\), giving \(x = \frac{y}{4}\).

Step 5: Set Integration Limits and Integrate

Integrate from \(y = 0\) to \(y = 4\). The volume integral becomes: \[ V = \pi \int_{0}^{4} \left[(\sqrt{y})^2 - \left(\frac{y}{4}\right)^2\right] \, dy \] Simplify the integrand to get: \[ V = \pi \int_{0}^{4} (y - \frac{y^2}{16}) \, dy \] Now, integrate term by term: \[ V = \pi \left[ \frac{y^2}{2} - \frac{y^3}{48} \right]_{0}^{4} \] Calculate the definite integral.

Step 6: Evaluate the Integral

Substitute \(y = 4\) and \(y = 0\) into the integrated result: \[ V = \pi \left[ \frac{4^2}{2} - \frac{4^3}{48} \right] - \pi \left[ \frac{0^2}{2} - \frac{0^3}{48} \right] \]\[ V = \pi \left[ 8 - \frac{64}{48} \right] = \pi \left[ 8 - \frac{4}{3} \right] = \pi \left[ \frac{24}{3} - \frac{4}{3} \right] = \pi \left[ \frac{20}{3} \right] \] Thus, \( V = \frac{20\pi}{3} \).

Key Concepts

Washer Method

The Washer Method is a powerful technique for finding the volume of a solid of revolution. When you rotate a region around an axis to form a 3D shape, such as a doughnut-like object, this method helps calculate the volume efficiently. Imagine slicing the solid into thin circular disks parallel to the axis of rotation. These disks resemble a washer, which is why the method is called the Washer Method. Each washer has an outer radius and an inner radius, and the volume of each is found by subtracting the volume of the smaller circle from the larger one.

• Outer Radius (\(R(y)\)): The larger circle's radius.
• Inner Radius (\(r(y)\)): The smaller circle's radius.

Given these radii, the formula for a washer is: \[ V = \pi\int_{y_{min}}^{y_{max}} \left( (R(y))^2 - (r(y))^2 \right) \, dy \] Understanding the washer setup and the relationship between radii is crucial for applying this method to find volumes of complex shapes.

Integration Limits

Integration limits define the starting and ending points for evaluating an integral. In the context of finding volumes using the Washer Method, these limits correspond to where the bounding curves intersect. In our exercise, two curves define the region that rotates around the axis:
- \(y = 4x\)
- \(y = x^2\)To locate where these curves meet, set them equal to each other: \(4x = x^2\). Solving this: \(x(x - 4) = 0\), gives two intersection points at \((0,0)\) and \((2,4)\). The y-values from these intersections determine the limits of integration, \(y = 0\) to \(y = 4\).

• These y-values come from solving for intersections in terms of x and then using those results to find the corresponding y-values.
• Using these points helps to set clear bounds for the integration process, ensuring the region is completely and accurately revolved.

Outer and Inner Radii

The outer and inner radii are fundamental to setting up the washer integral. They describe the distance from the axis of rotation to the outer and inner boundaries of the region that's being rotated. In the exercise, the radii depend on the other variables as follows:

• Outer Radius ({R(y)}): This is the distance from the axis of rotation to the further curve of the region. For the function \(y = x^2\), solve for \(x\) to get \(x = \sqrt{y}\). This expression forms the outer radius because it's the greater boundary in our region.
• Inner Radius ({r(y)}): This covers the space closer to the axis. For \(y = 4x\), solve for \(x\) to find \(x = \frac{y}{4}\). This expression dictates the inner boundary for the washers.

The outer and inner radii ensure that we're subtracting the inner shape from the outer shape correctly, capturing the region's true volume when revolved.

Definite Integrals

The definite integral computes the exact area, volume, or length for a specific range of values. In the washer method, the definite integral is used to add up an infinite number of infinitesimally thin washers' volumes that stretch from our lower to our upper limits of integration.
For our problem: \[ V = \pi \int_{0}^{4} \left((\sqrt{y})^2 - \left(\frac{y}{4}\right)^2\right) \, dy \]The integral simplifies to:\[ V = \pi \int_{0}^{4} \left(y - \frac{y^2}{16}\right) \, dy \]

• You evaluate this integral by solving each term separately, then substitute the limits to find the volume.
• Begin by integrating \(y\) and \(-\frac{y^2}{16}\) and substitute the bounds \(y = 0\) and \(y = 4\).

This results in a precise volume calculation, which in our case was \( V = \frac{20\pi}{3} \). Understanding how to evaluate definite integrals is key for finding exact values for rotating regions.