Problem 56 - Calculus of a Single Variable

Question

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

Step-by-Step Solution

Answer

The matched integral with the solid is: (i) Right Circular Cylinder, (ii) Right Circular Cone, (iii) Sphere, (iv) Ellipsoid, and (v) Torus.

Step 1: Associating solid geometrical shapes with the integrals

The geometrical solids are:- (a) A Right circular cone has a volume given by \( V=\frac{1}{3}\pi r^{2} h \)- (b) A Torus (or a donut shape) has a volume given by \( V=2\pi^{2}r^{2}R \)- (c) A Sphere has a volume given by \( V=\frac{4}{3}\pi r^{3} \)- (d) A Right circular cylinder has a volume given by \( V=\pi r^{2} h \)- (e) An Ellipsoid has a volume given by \( V=\frac{4}{3}\pi abc \)The integrals are:- (i) Most similarly represents the volume of cylinder, \( \left[\text{(d) Right circular cylinder} \right] \) because the volume of a cylinder is proportional to the radius \( r \) and the height \( h \). - (ii) Most similarly represents the volume of a right circular cone, \( \left[\text{(a) Right circular cone} \right] \), since the volume of this solid decreases linearly with the radius. - (iii) Most similarly represents the volume of a sphere, \( \left[\text{(c) Sphere} \right] \), because the term \( 2x \sqrt{r^{2}-x^{2}} \) signifies that the volume decreases as the square of \( r \) from the sphere's center to its surface.- (iv) Most similarly represents the volume of an ellipsoid, \( \left[\text{(e) Ellipsoid} \right] \), as \( a \) and \( b \) are involved with \( x \) in the integral.- (v) Most similarly represents the volume of a torus or doughnut shape, \( \left[\text{(b) Torus} \right] \), because it involves the radii \( r \) and \( R \) of the torus.

Step 2: Associating the dimensional parameters in each geometrical solid

Now let's understand the dimensions:- In a right circular cone, \( r \) is the base radius and \( h \) is the height. - In a torus, \( R \) is the distance from the center of the torus to the center of the tube and \( r \) is the radius of the tube.- In a sphere, \( r \) is the radius of the sphere.- In a right circular cylinder, \( r \) is the base radius and \( h \) is the height.- In an ellipsoid, \( a, b \) and \( c \) are the semi-axes.

Key Concepts

Right Circular Cone

A right circular cone is a three-dimensional shape featuring a circular base connected to a single vertex point above the base. To determine the volume of this cone, use the formula: \[ V = \frac{1}{3}\pi r^{2}h \]where:

• \(r\) is the radius of the base.
• \(h\) is the height from the base to the vertex.

This formula is derived by considering that the volume decreases linearly as you move from the base to the vertex. If you imagine slicing the cone horizontally, each slice is a circular disc, becoming smaller as you go up.
For integrals, we typically account for this changing disc size, which reflects in the integral setup similar to:\[ 2 \pi \int_{0}^{r} h x \left(1 - \frac{x}{r}\right) dx \]This integral represents the reduction of the cross-sectional area linearly with respect to \(x\) (distance from the base), thus closely related to the conical shape.

Torus

A torus, often likened to a donut, is essentially a circle revolved around an axis in its plane but not intersecting it. The volume formula for a torus is:\[ V = 2\pi^{2} r^{2} R \]where:

• \(R\) is the distance from the center of the tube to the center of the torus.
• \(r\) is the radius of the tube.

The volume calculation incorporates both rotational and radial symmetries. The integral representing the volume is set up based on slicing horizontally through the solid, where radii \( r \) and \( R \) determine the overall shape:\[ 2 \pi \int_{-r}^{r}(R-x) \left(2 \sqrt{r^{2}-x^{2}}\right) dx \]This reflects the nested circular paths embodied in the toroidal shape, combining major and minor radii in its volume calculation.

Sphere

A sphere is a perfectly symmetrical object, where every point on its surface is equidistant from its center. Its volume is calculated using the formula:\[ V = \frac{4}{3}\pi r^{3} \]where:

• \(r\) is the radius.

Understanding a sphere’s volume involves considering how the cross-sectional area changes as you move along its radius. Specifically, the volume integral is fashioned from repeatedly changing cross-sections as hemispherical slices:\[ 2 \pi \int_{0}^{r} 2 x \sqrt{r^{2}-x^{2}} dx \]This integral represents sweeping through half the sphere with discs, with \( x \) (the current slice position from the center) dictating the diameter of the discs, reflecting the spherical symmetry.

Ellipsoid Volumes

An ellipsoid resembles a stretched sphere, where the three axes of radius may differ. It is characterized by the volume formula:\[ V = \frac{4}{3}\pi abc \]where:

• \(a, b, c\) are the semi-principal axes.

The volume of an ellipsoid accounts for variations in its radii, applying a scaling to the sphere’s formula. This is reflected in the integral that adjusts each slice according to these axes:\[ 2 \pi \int_{0}^{b} 2 a x \sqrt{1-\frac{x^{2}}{b^{2}}} dx \]Here, the varying semi-axes describe the changing "slice thickness" across the ellipsoid, giving rise to a more nuanced calculation compared to a simple sphere, accommodating asymmetrical growth across its axes.