Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

The base of a solid is the region \(R\) bounded by \(y=\sqrt{x}\) and \(y=x^{2}\). Each cross section perpendicular to the \(x\)-axis is a semicircle with diameter extending across \(R\). Find the volume of the solid.

Step-by-Step Solution

Answer

The volume is \( \frac{9\pi}{560} \).

Step 1: Identify the Intersection Points

To find where the curves intersect, set the equations equal: \( \sqrt{x} = x^2 \). Squaring both sides gives \( x = x^4 \). Factoring out \( x \), we have \( x(x^3 - 1) = 0 \). Thus, \( x = 0 \) or \( x^3 - 1 = 0 \), which gives \( x = 1 \). The curves intersect at the points \((0,0)\) and \((1,1)\).

Step 2: Determine the Diameter of Semicircles

At a given \( x \), the diameter of the semicircle is the difference in the \( y \)-coordinates of the curves: \( \text{Diameter} = \sqrt{x} - x^2 \).

Step 3: Find the Area of Semicircles

The area \( A \) of a semicircle with diameter \( d \) is given by \( A = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi}{8}d^2 \). Substituting the diameter from step 2, the area is \( A = \frac{\pi}{8} (\sqrt{x} - x^2)^2 \).

Step 4: Set Up the Integral for Volume

The volume of the solid is the integral of the cross-sectional area along the \( x \)-axis from \( 0 \) to \( 1 \). So, \( V = \int_{0}^{1} \frac{\pi}{8} (\sqrt{x} - x^2)^2 \, dx \).

Step 5: Evaluate the Integral

Expand \((\sqrt{x} - x^2)^2 = x - 2x^{5/2} + x^4 \). The integral becomes \( V = \frac{\pi}{8} \int_{0}^{1} (x - 2x^{5/2} + x^4) \, dx \). Integrating gives \( \frac{\pi}{8} \left[ \frac{x^2}{2} - \frac{4}{7}x^{7/2} + \frac{x^5}{5} \right]_{0}^{1} = \frac{\pi}{8} \left( \frac{1}{2} - \frac{4}{7} + \frac{1}{5} \right) \).

Step 6: Simplify the Result

Calculate the expression inside the brackets: \( \frac{1}{2} = \frac{7}{14} \), \( -\frac{4}{7} = -\frac{8}{14} \), and \( \frac{1}{5} = \frac{2.8}{14} \). The sum is \( \frac{7}{14} - \frac{8}{14} + \frac{2.8}{14} = \frac{1.8}{14} \approx \frac{9}{70} \). Multiply by \( \frac{\pi}{8} \): \( V = \frac{\pi}{560} \times 9 = \frac{9\pi}{560} \).

Step 7: Final Step: Write the Volume

Thus the volume of the solid is \( \frac{9\pi}{560} \).

Key Concepts

Integral Calculus

Integral Calculus helps us find things like areas, volumes, central points, and many useful things. It's concerned with the whole picture, the accumulation of quantities, and working with continuous growth.
In this exercise, we used integration to calculate the volume of a solid. When we integrate, we're essentially summing up infinitely small pieces to find a total quantity.

• We find the cross-sectional area of the solid, which is a function of x, and integrates over the relevant interval.
• This process is done by calculating areas under curves and adding them together in a continuous manner.
• The integral provides us with a final accumulated total, which in our case, is the volume of the solid.

By using integral calculus, we transitioned from a cross-sectional view to understanding the full three-dimensional volume of the solid.

Volume of Solids

The volume of a solid can often be tricky to calculate, especially if the shape is not one of the basic geometric figures. In many calculus problems, we find the volume by integrating over a range.

• In our exercise, the solid has a base area defined by the intersection of two curves.
• The solid extends perpendicular to the base with defined cross-sections that are semicircles.
• By calculating the area of each cross-sectional slice and integrating it along the base, we compute the total volume.

The key is to correctly identify how your cross-sections fit into the shape, as each slice adds up to form the complete solid when the limits of integration are applied.

Cross-sectional Area

The cross-sectional area is a crucial concept when finding the volume of a solid. It refers to the area of an intersection that is sliced perpendicular to an axis, typically the x-axis for simplicity.
In this problem, the cross-sections are semicircles:

• First, determine the diameter of the semicircle, which is found by examining the space between the two bounding curves.
• This diameter can be expressed as a function in terms of x, which is given by the difference in y-values, or heights, of the curves.
• Using the formula for the area of semicircles, we substitute the diameter function and simplify it for integration.

The integration of these cross-sectional areas across the x-axis interval results in the total volume.

Intersection Points

Intersection points are where two curves meet or cross each other on a graph. These points are essential because they define the bounds for integration when dealing with areas and volumes.
To find the intersection points in this problem:

• Equate the expressions of the two curves, which correspond to the boundaries of the base region.
• Solve this equation to discover common solutions, which represent where the curves intersect each other.
• These solutions provide the limits of integration, x = 0 and x = 1, which are crucial for setting up the integral correctly.

Intersection points not only frame our area of interest but also help us in efficiently calculating cross-sections and their summation via integrals.