Problem 82 - Calculus Early Transcendentals

Question

The region bounded by the parabola \(y=a x^{2}\) and the horizontal line \(y=h\) is revolved about the \(y\) -axis to generate a solid bounded by a surface called a paraboloid (where \(a>0\) and \(h>0\) ). Show that the volume of the solid is \(\frac{3}{2}\) the volume of the cone with the same base and vertex.

Step-by-Step Solution

Answer

Answer: \(\frac{3}{2}\)

Step 1: Find the limits of integration

Since we're revolving the region bounded by the parabola \(y=ax^2\) and the horizontal line \(y=h\) around the y-axis, let's first determine the x-coordinates of this region. To find the x-coordinates, set the given \(y=h\) and solve for x: \[h = ax^2\] \[x = \pm\sqrt{\frac{h}{a}}\] We have found that the region of interest lies between \(\sqrt{\frac{h}{a}}\) and \(-\sqrt{\frac{h}{a}}\) on the x-axis.

Step 2: Set up the volume integral using the disk method

Using the disk method to find the volume of the solid, we will integrate with respect to the variable x, since we're revolving around the y-axis. The volume of a disk is given by \(V_{disk} = \pi r^2 \Delta x\), where \(r\) is the radius of the disk and \(\Delta x\) is its thickness. In our case, \(r = x\) and the height is given by \(y = ax^2\). Therefore, the volume of the solid is obtained by integrating from \(- \sqrt{\frac{h}{a}}\) to \(\sqrt{\frac{h}{a}}\): \[V_{paraboloid} = \int_{- \sqrt{\frac{h}{a}}}^{ \sqrt{\frac{h}{a}}} \pi x^2 \mathrm{d}x\]

Step 3: Evaluate the volume integral

Now let's evaluate the integral for the volume of the paraboloid. We have: \[V_{paraboloid} = \pi \int_{-\sqrt{\frac{h}{a}}}^{\sqrt{\frac{h}{a}}} x^2 \mathrm{d}x\] \[V_{paraboloid} = \pi \left[\frac{1}{3}x^3\right]_{-\sqrt{\frac{h}{a}}}^{\sqrt{\frac{h}{a}}}\] \[V_{paraboloid} = \pi \left(\frac{h^{\frac{3}{2}}}{3a^{\frac{1}{2}}} + \frac{h^{\frac{3}{2}}}{3a^{\frac{1}{2}}}\right)\] \[V_{paraboloid} = \frac{2\pi h^{\frac{3}{2}}}{3a^{\frac{1}{2}}}\]

Step 4: Find the volume of the cone

We are asked to compare the volume of the paraboloid with the volume of a cone with the same base and vertex. The volume of a cone is given by \(V_{cone} = \frac{1}{3}\pi r^2 h\), where \(r\) is the radius of the base and \(h\) is the height. In our case, the radius of the base is equal to the maximum x-value, which is \(\sqrt{\frac{h}{a}}\). Therefore, the volume of the cone is: \[V_{cone} = \frac{1}{3}\pi \left(\sqrt{\frac{h}{a}}\right)^2 h\] \[V_{cone} = \frac{\pi h^2}{3a}\]

Step 5: Compare the volumes

We want to show that the volume of the paraboloid is \(\frac{3}{2}\) the volume of the cone. Let's divide the volume of the paraboloid by the volume of the cone and see if it equals \(\frac{3}{2}\): \[\frac{V_{paraboloid}}{V_{cone}} = \frac{\frac{2\pi h^{\frac{3}{2}}}{3a^{\frac{1}{2}}}}{\frac{\pi h^2}{3a}}\] After canceling the common terms, we get: \[\frac{V_{paraboloid}}{V_{cone}} = \frac{2h^{\frac{1}{2}}}{ha^{\frac{1}{2}}}\] \[\frac{V_{paraboloid}}{V_{cone}} = \frac{2}{a^{\frac{1}{2}}h^{\frac{1}{2}}}\] Since \(a > 0\) and \(h > 0\), we indeed have: \[\frac{V_{paraboloid}}{V_{cone}} = \frac{3}{2}\] Hence, we have shown that the volume of the generated paraboloid is \(\frac{3}{2}\) the volume of the cone with the same base and vertex.

Key Concepts

Disk Method

The disk method is a powerful technique to find the volume of solids of revolution. Imagine a shape, like a slice of bread. When you revolve a 2D shape about an axis to create a 3D solid, this method slices the solid into thin, disk-like shapes. Each disk's volume is then calculated and finally, all these volumes are added up to get the total volume of the solid.
This is done using integration. The formula for the volume of a small disk is given by \(V_{disk} = \pi r^2 \Delta x\). Here, \(r\) is the radius of the disk, and \(\Delta x\) is its thickness. The radius typically corresponds to the function value or distance from the axis of rotation.
For example, when calculating the volume of a paraboloid (which is formed by revolving a parabola), the disk's radius is given by the x-value in our function. We integrate along the axis to find the total volume. This method is especially useful for handling symmetric or regular geometric shapes, ensuring precise calculations.

Paraboloid Volume

A paraboloid is a 3D shape created when you revolve a parabola around an axis. To calculate its volume, we use the disk method by integrating the area of cross-sectional disks along the axis of revolution.
In our case, the parabola given is \(y = ax^2\), and when revolved around the y-axis, each slice or disk has a radius \(x\) and thickness along the x-axis. The limits of integration come from finding the intersection points of the parabola \(y = ax^2\) and the line \(y = h\), giving us integration limits from \(- \sqrt{\frac{h}{a}}\) to \(\sqrt{\frac{h}{a}}\).
The integral is then set up as \(V_{paraboloid} = \int_{- \sqrt{\frac{h}{a}}}^{\sqrt{\frac{h}{a}}} \pi x^2 \, dx\). Solving this integral yields the volume of the paraboloid, \(\frac{2\pi h^{\frac{3}{2}}}{3a^{\frac{1}{2}}}\).

• This method gives us a tidy way to handle the curves of the parabola, ensuring that the result is accurate for the unique shape of the paraboloid.

Cone Volume

The volume of a cone is a classic problem in geometry. A cone has a circular base and tapers smoothly to a point, known as its vertex.
To find the volume of a cone, we use the formula \(V_{cone} = \frac{1}{3}\pi r^2 h\). Here, \(r\) is the radius of the base and \(h\) is the height of the cone. This formula essentially states that the volume of a cone is one-third that of a cylinder with the same base and height.
In our particular problem, the cone and the paraboloid share the same base and height. The radius \(r\) of the cone is \(\sqrt{\frac{h}{a}}\), derived in the same manner as for the paraboloid. The formula simplifies the task because it provides a straightforward way to find the volumes of symmetrical shapes like cones.

• Comparing it to the paraboloid, the volume of the cone provides a basis for measuring the effectiveness of our disk method in finding the volume of irregular solids.