Question

Match each integral with the solid whose volume it represents, and give the dimensions of each solid. (a) Right circular cylinder (b) Ellipsoid (c) Sphere (d) Right circular cone (e) Torus (i) \(\pi \int_{0}^{h}\left(\frac{r x}{h}\right)^{2} d x\) (ii) \(\pi \int_{0}^{h} r^{2} d x\) (iii) \(\pi \int_{-r}^{r}\left(\sqrt{r^{2}-x^{2}}\right)^{2} d x\) (iv) \(\pi \int_{-b}^{b}\left(a \sqrt{1-\frac{x^{2}}{b^{2}}}\right)^{2} d x\) (v) \(\pi \int_{-r}^{r}\left[\left(R+\sqrt{r^{2}-x^{2}}\right)^{2}-\left(R-\sqrt{r^{2}-x^{2}}\right)^{2}\right] d x\)

Short answer

(a) Right circular cylinder -> (ii) \(\pi \int_{0}^{h} r^{2} d x\), (b) Ellipsoid -> (iv) \(\pi \int_{-b}^{b}\left(a \sqrt{1-\frac{x^{2}}{b^{2}}}\right)^{2} d x\), (c) Sphere -> (iii) \(\pi \int_{-r}^{r}\left(\sqrt{r^{2}-x^{2}}\right)^{2} d x\), (d) Right circular cone -> (i) \(\pi \int_{0}^{h}\left(\frac{r x}{h}\right)^{2} d x\), (e) Torus -> (v) \(\pi \int_{-r}^{r}\left[\left(R+\sqrt{r^{2}-x^{2}}\right)^{2}-\left(R-\sqrt{r^{2}-x^{2}}\right)^{2}\right] d x\)

Step-by-step solution

Identify volume formulas

To solve this exercise, all known volume formulae for solids such as cones, spheres, ellipsoids, cylinders, and tori need to be understood and identified.

Match integrals with solids

Next, each volume formula needs to be matched with the corresponding integral that represents it.

Detail the matches

(a) Right circular cylinder has the volume formula \(V = \pi r^{2}h\), which corresponds to integral (ii).\ (b) Ellipsoid has the volume formula \(V = \frac{4}{3}\pi ab^2\), which corresponds to integral (iv).\ (c) Sphere has the volume formula \(V = \frac{4}{3}\pi r^{3}\), which corresponds to integral (iii).\ (d) Right circular cone has the volume formula \(V = \frac{1}{3}\pi r^{2}h\), which corresponds to integral (i).\ (e) Torus (doughnut shape) does not have a standard volume formula, but it corresponds to integral (v).

Key concepts

Understanding the Volume of a Right Circular Cylinder

The volume of a right circular cylinder can be easily understood by considering its two main parts: the circular base with radius 'r' and the height 'h'. Imagine stacking multiple circles, one on top of the other, creating a height 'h'. The formula for the volume of a cylinder is expressed as \( V = \pi r^2 h \).

To grasp this concept, picture a soda can: its volume is the area of the circle (the base) multiplied by the can's height. In calculus, this volume can be found using the integral \( \pi \int_{0}^{h} r^2 dx \), which represents the sum of infinitesimally thin circular disks, each with the area \( \pi r^2 \), stacked from the bottom to the top of the cylinder.

Calculating Ellipsoid Volume

An ellipsoid resembles a stretched or compressed sphere, often looking like an oval-shaped ball. Its volume is more complex to visualize than that of a cylinder or sphere. Nevertheless, it's given by the formula \( V = \frac{4}{3}\pi abc \), where 'a', 'b', and 'c' are the semi-axes lengths in each of the three dimensions.

Intuitively, you can think of the ellipsoid volume as taking a sphere's volume formula and adjusting it for each directional stretch. The integral that matches this is \( \pi \int_{-b}^{b}(a \sqrt{1-\frac{x^2}{b^2}})^2 dx \), incorporating the different stretches of the semi-axes into the calculus.

Sphere Volume Insights

The volume of a sphere, one of the most symmetrical solids, is derived from its radius 'r'. The formula \( V = \frac{4}{3}\pi r^3 \) shows that the volume depends entirely on the radius, encapsulating the concept of three-dimensional symmetry around a central point.

To understand the corresponding integral \( \pi \int_{-r}^{r}(\sqrt{r^2 - x^2})^2 dx \), think of it as summing up the areas of infinitely many concentric circles (each representing a cross-section of the sphere) from one end of the sphere to the other.

Right Circular Cone Volume Comprehension

The right circular cone, reminiscent of an ice cream cone, tapers smoothly from a circular base to a point. Its volume involves both the base radius 'r' and the height 'h' and is computed using the formula \( V = \frac{1}{3}\pi r^2 h \). This signifies that the cone's volume is a third of what a cylinder with the same base and height would have.

In calculus terms, the integral \( \pi \int_{0}^{h}(\frac{r x}{h})^2 dx \) symbolizes the sum of ever-decreasing circles as one moves from the base to the point, showing the linear decrease in radius as you go along the height.

Exploring Torus Volume

A torus is a donut-shaped solid formed by revolving a small circle around an axis outside of the circle. It's described not just by one radius, but by two: 'r', the radius of the generating circle, and 'R', the distance from the center of the torus to the center of the tube.

Understanding its volume through integration can be tricky, but the corresponding integral \( \pi \int_{-r}^{r}[(R + \sqrt{r^2 - x^2})^2 - (R - \sqrt{r^2 - x^2})^2] dx \) accounts for the complex shape by calculating the volume of a series of concentric, rotating circles. It requires considering both the inner void and the material thickness, accurately depicting the unique volume of a torus.