Problem 40 - Single Variable Calculus

Question

In Exercises 37-42, find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated line. $$ y=x^{2}, \quad y=\frac{1}{2} x^{2}+2 ; \quad \text { the line } y=5 $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the graphs of the equations \(y=x^2\) and \(y=\frac{1}{2}x^2+2\) about the line \(y=5\) is approximately \(22.619 \pi\).

Step 1: Find the points of intersection

To find the points of intersection between the functions \(y=x^2\) and \(y=\frac{1}{2}x^2 + 2\), we set them equal to each other: \[x^2 = \frac{1}{2}x^2 + 2\] Subtract \(\frac{1}{2}x^2\) from both sides: \[\frac{1}{2}x^2 = 2\] Multiply both sides by 2: \[x^2 = 4\] Take the square root of both sides: \[x = \pm 2\] The two functions intersect at \(x = -2\) and \(x = 2\). These are the limits of integration.

Step 2: Set up the washer method integral

To set up the washer method integral, we need to calculate the radii of the outer and inner washers formed by the region bounded by the functions. Outer radius \(R(x)\): The distance between the line \(y = 5\) and the function \(y = x^2\) is given by: \[R(x) = 5 - x^2\] Inner radius \(r(x)\): The distance between the line \(y = 5\) and the function \(y = \frac{1}{2}x^2 + 2\) is given by: \[r(x) = 5 - (\frac{1}{2}x^2 + 2)\] Now we will set up the integral representing the volume: \[V = \pi \int_{-2}^2 \left[R(x)^2 - r(x)^2\right] dx\] \[V = \pi \int_{-2}^2 \left[(5-x^2)^2 - \left(5-\frac{1}{2}x^2 - 2\right)^2\right] dx\]

Step 3: Evaluate the integral

Now, let's calculate the integral: \[V = \pi \int_{-2}^2 \left[(5-x^2)^2 - \left(5-\frac{1}{2}x^2 - 2\right)^2\right] dx\] \[V = \pi \int_{-2}^2 \left[(25 - 10x^2 + x^4) - \left(\frac{1}{4}x^4 - x^3 + \frac{15}{2}x^2 - 10x + 9\right)\right] dx\] \[V = \pi \int_{-2}^2 \left[\frac{3}{4}x^4 - x^3 + \frac{5}{2}x^2 - 10x + 16\right] dx\] Next, we will integrate each term with respect to x: \[V = \pi \left[\frac{3}{20}x^5 - \frac{1}{4}x^4 + \frac{5}{6}x^3 - 5x^2 + 16x\right]_{-2}^2\] Finally, we will evaluate the definite integral using the limits of integration: \[V = \pi \left(\left[\frac{3}{20}(2)^5 - \frac{1}{4}(2)^4 + \frac{5}{6}(2)^3 - 5(2)^2 + 16(2)\right] - \left[\frac{3}{20}(-2)^5 - \frac{1}{4}(-2)^4 + \frac{5}{6}(-2)^3 - 5(-2)^2 + 16(-2)\right]\right)\] \[V = \pi (7.2)\] Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line \(y = 5\) is \(\approx 22.619 \pi\).

Key Concepts

Washer Method

The Washer Method is a technique used to find the volume of a solid of revolution. Imagine slicing a shape into thin, circular disks (washers). You can then calculate the volume of each washer and add them up. This method is perfect when you have two functions forming an outer and inner boundary.

For this problem, we revolve the region defined by the functions about the line \(y = 5\). Each washer here has an outer radius \(R(x) = 5 - x^2\) and an inner radius \(r(x) = 5 - \left(\frac{1}{2}x^2 + 2\right)\). These radii form the difference squared in the integral, representing each washer's volume.

Calculating the volume involves setting up the integral:

• Subtract the square of the inner radius from the square of the outer radius.
• Integrate this difference from the limits of intersection.

By evaluating this integral, you determine the total volume of the solid.

Definite Integral

A Definite Integral calculates the area under a curve within specified bounds. When applied to volumes of solids, it helps sum the volumes of infinitesimally thin slices or washers.

Given the problem, the integral is expressed as follows:
\[V = \pi \int_{-2}^{2} \left[(5-x^2)^2 - \left(5 - \frac{1}{2}x^2 - 2\right)^2\right] dx\]

This integral represents the volume of the revolved region. Here, we integrate from \(-2\) to \(2\), capturing the total volume across these boundaries.

The definite integral provides the exact volume, as it sums all infinitesimal slices or washer volumes between the limits efficiently.

Points of Intersection

Finding Points of Intersection is vital in setting up your integral’s boundaries. These are the places where your functions meet, marked here by solving:
\[x^2 = \frac{1}{2}x^2 + 2\]

Solving gives \(x = -2\) and \(x = 2\). These intersections are crucial as they define the limits for integration. They help capture the region between curves, ensuring that the washer method encapsulates exact boundaries.

Visually, think of these points as the start and end of the area you’re revolving to form a three-dimensional shape.

Limits of Integration

Limits of Integration are the bounds between which you evaluate your integral. They are critical in defining where the region of interest starts and ends.

For the exercise, the limits are \(-2\) and \(2\), found through the points of intersection. These limits ensure that we only consider the overlapping segments of the curves

By establishing clear limits, the calculation uses only the relevant part of the graphical region, ensuring accuracy in the final volume determination.