Exercises: General Slicing

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Exercises computing volume by cross-sectional area.

The base of a solid region is bounded by the curves $x = 0$ , $y = x^2$ and $y = x$ . The cross sections perpendicular to the $x$ -axis are squares. Compute the volume of the region.

A typical square cross section has side length $L = \answer {x - x^2}$ and area $A = \answer {(x-x^2)^2}$ .
Possible numerical values of the $x$ -coordinates of points in the base range from a minimum value of $x = \answer {0}$ up to a maximum of $x = \answer {1}$ .
To compute volume, integrate: $V = \int _{\answer {0}}^{\answer {1}} \answer {(x-x^2)^2} d\answer {x} = \answer {\frac {1}{30}}.$

Find the volume of the region in three-dimensional space defined by the inequalities $\begin{align*} 0 & \leq x \leq 1, \\ 0 & \leq y \leq z^2, \\ 0 & \leq z \leq 3. \end{align*}$

Cross sections perpendicular to the $z$ -axis are with length $\answer {1}$ in the $x$ -direction and width $\answer {z^2}$ in the $y$ -direction.
The area of a $z$ cross section is $A(z) = \answer {z^2}$ .
To compute volume, integrate: $V = \int _{\answer {0}}^{\answer {3}} \answer {z^2} dz = \answer {9}.$

A right circular cylinder of radius $1$ and height $3$ is twisted along its axis so that the disk at height $z$ is centered on the axis $x = \cos (2 \pi z/3), y = \sin (2 \pi z/3))$ , which corresponds to one full twist along the axis. Compute the volume of this twisted cylinder. $V = \answer {3 \pi }.$

A certain three-dimensional region has a base in the $xy$ -plane which is bounded above by the graph $y = 1-x^2$ and below by $y=0$ . Slices perpendicular to the $y$ -axis are equilateral triangles whose base lies in the $xy$ -plane as well. Compute the volume of the region. $V = \answer {\frac {\sqrt {3}}{2}}.$

Sample Exam Questions

(2018 Midterm 1) Compute the volume of the region in 3-dimensional space which satisfies the inequalities $\begin{align*} 0 \leq x \leq (1 - z^2) \qquad \mbox{ and } \qquad 0 \leq y \leq (1 + z^2) \qquad \mbox{ and } \qquad 0 \leq z \leq 1. \end{align*}$

$\displaystyle \frac {2}{3}$ $\displaystyle \frac {3}{4}$ $\displaystyle \frac {4}{5}$ $\displaystyle \frac {5}{6}$ $\displaystyle \frac {6}{7}$ none of these

(2019 Midterm 1) The inequality $\frac {x^2}{a^2} + \frac {y^2}{b^2} \leq 1$ defines an ellipse in the $xy$ -plane whose area is $\pi a b$ for any positive values of the constants $a$ and $b$ . Compute the three dimensional volume of the region defined by $4 x^2 + z^2 y^2 \leq z^2 \mbox { for } 0 \leq z \leq 1.$ (Hints won’t reveal until after you choose a response.)

$\displaystyle 4 \pi z^2$ $4 \pi$ $\displaystyle \frac {\pi }{4z}$ $\displaystyle \frac {\pi }{4}$ $\displaystyle {\pi z}$ $\displaystyle {\pi }$

Dividing the first inequality by $z^2$ on both sides gives $\frac {x^2}{\frac {z^2}{4}} + \frac {y^2}{1} \leq 1,$

which means that slices in the $z$ -direction are ellipses with area $\pi \frac {z}{2} \cdot 1 = \frac {\pi z}{2}$ .

Volume is obtained by integration: $V = \int _0^1 \frac {\pi z}{2} dz = \left . \frac {\pi z^2}{4} \right |_{0}^1 = \frac {\pi }{4}.$