6.10E: Exercises

Reading Questions

1. What is the definition of a solid's cross-section?
2. What is the general definite integral formula for the volume of a solid using the slicing method, where \(A(x)\) is the cross-sectional area at \(x\)?
3. List the three steps in the problem-solving strategy for finding volumes by the slicing method.
4. In Example 1.2.1, what geometric principle is used to find the side length \(s\) of the square cross-section in terms of its distance \(x\) from the apex of the pyramid?
5. If the cross-sections of a solid are taken perpendicular to the y-axis, what variable must the area function and the integration be in terms of?
6. According to the text, what is the mathematical definition of a cylinder? Does its base have to be a circle?

1) Derive the formula for the volume of a sphere using the slicing method.

2) Use the slicing method to derive the formula for the volume of a cone.

3) Use the slicing method to derive the formula for the volume of a tetrahedron with side length \(a.\)

Volumes by Slicing

For exercises 4 - 8, draw a typical slice and find the volume using the slicing method for the given volume.

4) A pyramid with height 6 units and square base of side 2 units, as pictured here.

Solution:

Here the cross-sections are squares taken perpendicular to the \(y\)-axis.
We use the vertical cross-section of the pyramid through its center to obtain an equation relating \(x\) and \(y\).
Here this would be the equation, \( y = 6 - 6x \). Since we need the dimensions of the square at each \(y\)-level, we solve this equation for \(x\) to get, \(x = 1 - \tfrac{y}{6}\).
This is half the distance across the square cross-section at the \(y\)-level, so the side length of the square cross-section is, \(s = 2\left(1 - \tfrac{y}{6}\right).\)
Thus, we have the area of a cross-section is,

\(A(y) = \left[2\left(1 - \tfrac{y}{6}\right)\right]^2 = 4\left(1 - \tfrac{y}{6}\right)^2.\)

\(\begin{align*} \text{Then},\quad V &= \int_0^6 4\left(1 - \tfrac{y}{6}\right)^2 \, dy \\[5pt]
&= -24 \int_1^0 u^2 \, du, \quad \text{where} \, u = 1 - \tfrac{y}{6}, \, \text{so} \, du = -\tfrac{1}{6}\,dy, \quad \implies \quad -6\,du = dy \\[5pt]
&= 24 \int_0^1 u^2 \, du = 24\dfrac{u^3}{3}\bigg|_0^1 \\[5pt]
&= 8u^3\bigg|_0^1 \\[5pt]
&= 8\left( 1^3 - 0^3 \right) \quad= \quad 8\, \text{units}^3 \end{align*}\)

5) A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

6) A tetrahedron with a base side of 4 units,as seen here.

Answer

\(V = \frac{32}{3\sqrt{2}} = \frac{16\sqrt{2}}{3}\) units³

7) A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

8) A cone of radius \( r\) and height \( h\) has a smaller cone of radius \( r/2\) and height \( h/2\) removed from the top, as seen here. The resulting solid is called a frustum.

Answer

\(V = \frac{7\pi}{12} hr^2\) units³

For exercises 9 - 14, draw an outline of the solid and find the volume using the slicing method.

9) The base is a circle of radius \( a\). The slices perpendicular to the base are squares.

10) The base is a triangle with vertices \( (0,0),(1,0),\) and \( (0,1)\). Slices perpendicular to the \(xy\)-plane are semicircles.

Answer

\( V = \int_0^1 \frac{\pi(1-x)^2}{8}\, dx \quad = \quad \frac{ \pi }{24}\) units³

11) The base is the region under the parabola \( y=1−x^2\) in the first quadrant. Slices perpendicular to the \(xy\)-plane are squares.

12) The base is the region under the parabola \( y=1−x^2\) and above the \(x\)-axis. Slices perpendicular to the \(y\)-axis are squares.

Answer

\( V = \int_0^1 4(1 - y)\,dy \quad = \quad 2\) units³

13) The base is the region enclosed by \( y=x^2\) and \( y=9.\) Slices perpendicular to the \(x\)-axis are right isosceles triangles.

14) The base is the area between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

Answer

\( V = \int_0^1 \frac{\pi}{8}\left( x - x^2 \right)^2 \, dx \quad=\quad \frac{ \pi }{240}\) units³

For exercises 15 - 16, find the volume of the solid described.

15) The base is the region between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

16) The base is the region enclosed by the generic ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\) Slices perpendicular to the \(x\)-axis are semicircles.

Answer

\(V = \frac{2ab^2 \pi }{3}\) units³

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.