Problem 40 - Calculus of a Single Variable

Question

Finding the Volume of a Solid In Exercises \(37-40,\) use the integration capabilities of a graphing utility to approximate the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=\sqrt{2 x}, \quad y=x^{2} $$

Step-by-Step Solution

Answer

The volume of the solid generated is \( V = \pi [2 - \frac{2\sqrt{2}}{5}]\) cubic units.

Step 1: Graph the Functions and Find the Points of Intersection

First, plot the equations \(y=\sqrt{2x}\) and \(y=x^2\) using a graphing tool. Observe their points of intersection. To find the exact points of intersection, set the two equations equal to each other: \(\sqrt{2x} = x^2\). That simplifies to \(2x = x^4\). Factoring, one finds \(x=0\) and \(x=\sqrt{2}\). Those will be the limits of the integral.

Step 2: Setup the Integral

The volume \(V\) of the solid is given by the formula \(V=\pi \int_{a}^{b} [f(x)]^2 \, dx\), where \(f(x)\) is the outer radius function, \(a\) and \(b\) are the limits of integration. Here \(f(x) = \sqrt{2x}\) is the outer radius, as it is greater than \(x^2\) on the interval [0, \(\sqrt{2}\)]. The integral becomes \(V = \pi \int_{0}^{\sqrt{2}} (\sqrt{2x})^2 - (x^2)^2 dx\). This simplifies to \(V = \pi \int_{0}^{\sqrt{2}} [2x - x^4] dx\).

Step 3: Evaluate the Integral

Next, the integral may be evaluated. The antiderivative of \(2x - x^4\) is \(x^2 - \frac{1}{5}x^5 + C\). The definite integral from \(0\) to \(\sqrt{2}\) can be found using the Fundamental Theorem of Calculus, which states that \( \int_{a}^{b} f(x) dx = F(b) - F(a)\), where \(F(x)\) is the antiderivative of f. The volume is therefore given by \(V = \pi [(\sqrt{2})^2 - \frac{1}{5}(\sqrt{2})^5 - (0)^2 + \frac{1}{5}(0)^5] = \pi[2 - \frac{2\sqrt{2}}{5}]\).

Key Concepts

Integration to Find Volume

When you encounter a problem involving finding the volume of a solid shape, integration comes as a powerful tool. This is particularly true when the shape is complex or irregular, such as when it's created by revolving a function around an axis. Imagine cutting the solid into infinitely thin cross-sections perpendicular to the axis of revolution; integration effectively sums up the volume of these slices.

The general formula to find the volume of a solid of revolution around the x-axis uses integration in the form \( V = \pi \int_{a}^{b} [f(x)]^2 dx \), where \( f(x) \) is the radius of the solid at any point \( x \), and \( a \) and \( b \) are the bounds of the shape along the x-axis. Through integration, you'll create a series of disks with radii \( f(x) \) and infinitesimally small thickness, whose collective volume comprises the volume of the solid.

Graphing Utility Application

A graphing utility is an invaluable asset when approaching calculus problems, especially when visualizing the problem before solving. To understand the solid you will be working with, graph each function and identify their intersection points, which will serve as the limits for your volume calculation.

Using these tools, you can visualize the area that will revolve around an axis. In our case, the area between \( y=\sqrt{2x} \) and \( y=x^2 \) will be rotated around the x-axis. This visual aid can prevent mistakes in setting up your integral as it clearly denotes which function forms the outer radius and which the inner, if applicable.

Solids Generated by Revolution

Solids generated by revolution are unique three-dimensional shapes created by rotating a two-dimensional area around a line (the axis of revolution). Think of it like a potter's clay spinning on a wheel, the hands of the potter press into the clay, creating curves and contours that are symmetric around the center.

In calculus, you'll often need to find the volume of a shape formed by revolving a region, called the washer or disk depending on the presence of a hole, around an axis. If the area does not enclose the axis of rotation, you'll generate a solid resembling a ring - which is where the 'washer method' comes in, utilizing the outer and inner radii of these ring-shaped cross-sections to calculate volume.

Volume by Integration

The final step is evaluating the integral to obtain the volume. For our exercise, once the equation \( V = \pi \int_{0}^{\sqrt{2}} (2x - x^4) dx \) is set up, we calculate the antiderivative and apply the limits using the Fundamental Theorem of Calculus. This theorem connects differentiation with integration, letting you evaluate the integral by simply substituting the upper and lower bounds into the antiderivative.

Completing the integration process grants you the exact volume of the solid. Remember, integration isn't just about finding areas under curves; it's just as vital for calculating volumes of complex shapes in three dimensions, showcasing the real-world applicability of calculus.