Question

The base of a solid is the region bounded by \(y=1-x^{2}\) and \(y=1-x^{4}\). Cross sections of the solid that are perpendicular to the \(x\) -axis are squares. Find the volume of the solid.

Short answer

The volume of the solid is \(\frac{16}{315}\) cubic units.

Step-by-step solution

Determine the Intersection Points

To find the points where the boundaries intersect, we set the equations equal: \[ 1-x^2 = 1-x^4 \] Solving for \(x\), first cancel \(1\):\[ -x^2 = -x^4 \]Rearrange and factor:\[ x^4 - x^2 = 0 \]\[ x^2(x^2 - 1) = 0 \]Thus, \(x^2 = 0\) or \(x^2 = 1\). Therefore, \(x = 0, \pm 1\). The intersections are at \(x = -1, 0, 1\).

Find the Height of the Square Cross-sectional Area

The height of the square cross-section at any point \(x\) is the difference of the two functions:\[ y = 1-x^2 \text{ and } y = 1-x^4 \].The side length of the square is:\[ s = (1-x^2) - (1-x^4) = x^4 - x^2 \].

Set Up the Integral for Volume

The volume \(V\) of the solid can be computed by integrating the area of each square (\(s^2\)) from \(x = -1\) to \(x = 1\):\[ V = \int_{-1}^{1} (x^4 - x^2)^2 \, dx \].Expand the square:\[ (x^4 - x^2)^2 = x^8 - 2x^6 + x^4 \].

Evaluate the Integral

Calculate the integral:\[ V = \int_{-1}^{1} (x^8 - 2x^6 + x^4) \, dx \]\[ = \int_{-1}^{1} x^8 \, dx - 2\int_{-1}^{1} x^6 \, dx + \int_{-1}^{1} x^4 \, dx \].Since all are even functions, they're symmetric about the origin:\[ 2 \times \left( \int_{0}^{1} x^8 \, dx - 2\int_{0}^{1} x^6 \, dx + \int_{0}^{1} x^4 \, dx \right) \].

Compute Each Sub-Integral

Compute each part:\[ \int_{0}^{1} x^8 \, dx = \frac{1}{9}x^9 \bigg|_{0}^{1} = \frac{1}{9} \]\[ \int_{0}^{1} x^6 \, dx = \frac{1}{7}x^7 \bigg|_{0}^{1} = \frac{1}{7} \]\[ \int_{0}^{1} x^4 \, dx = \frac{1}{5}x^5 \bigg|_{0}^{1} = \frac{1}{5} \].

Calculate the Final Volume

Substitute back:\[ V = 2 \left( \frac{1}{9} - 2\left( \frac{1}{7} \right) + \frac{1}{5} \right) \]\[ = 2 \left( \frac{1}{9} + \frac{1}{5} - \frac{2}{7} \right) \]Find common denominator (315):\[ = 2 \left( \frac{35}{315} + \frac{63}{315} - \frac{90}{315} \right) \]\[ = 2 \left( \frac{8}{315} \right) \]\[ = \frac{16}{315} \].

Final Answer

The volume of the solid is \(\frac{16}{315}\) cubic units.

Key concepts

Volume of Solids

Understanding the volume of solids is a fundamental concept in calculus and geometry. This involves calculating the space that a solid object occupies in three-dimensional space.

When a solid has a known shape, finding its volume can be straightforward, as with cubes or spheres. However, in calculus, we often deal with irregularly shaped solids, which require more advanced techniques to compute volume.

For instance, when the cross-sections of a three-dimensional solid are known, we can find its volume by integrating the area of these cross-sections along the axis of interest. This is particularly useful when the cross-sections change size or shape depending on their position along this axis, as with our example where cross sections are squares with varying side lengths. This technique is powerful because it allows us to compute the volume of complex shapes by breaking them into simpler, uniform components.

Integral Calculus

Integral calculus provides the tools to compute quantities like area, volume, and other values that accumulate over a range. In a nutshell, this branch of calculus deals with the inverse process of differentiation, which is finding a function given its rate of change.

When it comes to calculating the volume of a solid, integral calculus allows us to sum an infinite number of infinitely small cross-sectional areas. Essentially, we calculate the area of each thin slice of the solid, and integrate over the entire range where these slices are located.

In our exercise, setting up the integral involves first determining the formula for the cross-sectional area at any point and then integrating this formula over the range of interest. The integral not only allows us to deduce the formula but also provides the computational framework to find the total volume efficiently.

Cross-Sectional Area

The concept of cross-sectional area is pivotal when dealing with volumes of solids in calculus. It involves slicing a solid perpendicular to a specified axis and analyzing the shape and area of these slices.

In our problem, cross-sectional areas are squares. For any given point along the x-axis, the side length of each square was determined by the difference in the y-values of two given functions. This difference defined how large each cross-section would be.

The formula for each cross-sectional area could be determined by knowing the geometry of these slices—in this case, squares. The area of a square is simply the square of the side length. Thus, the variance in side length as dictated by the functions helps define the shape of the overall solid.

Using these cross-sectional areas, integral calculus then enables us to assemble these areas into a complete volume measure, giving insights into the total space occupied by the solid.