Assignment #2 : Volumes

Geometric Derivation of the Volume of a Cone

The volume of a cone with base radius $r$ and height $h$ is given by the standard formula:

$V = \frac{1}{3}\pi r^2 h$

Conceptual Analysis of Fallacious Formulations

Attempting to calculate the volume of a cone directly using the simple cylindrical volume formula $V = \pi r^2 h$ leads to significant error. The equation $V = \pi r^2 h$ assumes that every horizontal cross-section along the height $h$ possesses a constant radius equal to the maximum base radius $r$. This describes a right circular cylinder.

In a cone, the cross-sectional radius is non-constant; it varies continuously as a linear function of distance from the apex, ranging from $0$ at the tip to $r$ at the base. Substituting the fixed maximum radius $r$ into the uniform cylinder formula treats the entire figure as a cylinder of uniform radius, thereby overestimating the volume by a factor of $3$.

Coordinate Setup and Disk Slicing Geometry

To derive the exact volume using calculus, position the cone on a Cartesian coordinate plane:

1. Place the apex/tip of the cone at the origin $(0,0)$.

2. Direct the central axis of symmetry along the positive $x$-axis, extending from $x = 0$ to $x = h$

3. The base of the cone sits vertically at $x = h$ with a maximum radius of $r$.

4. The generator line outlining the upper boundary of the cone's cross-section passes through $(0,0)$ and $(h, r)$. The equation of this bounding line is:

$y(x) = \frac{r}{h}x$

5. Slice the cone perpendicular to the $x$-axis into $n$ narrow cylindrical disks. Keeping the circular cross-sections intact ensures straightforward area computation.

Differential Disk Volume Expression

Each cylindrical disk slice located at position $x$ has:

• Radius equal to the vertical height of the line: $y(x) = \frac{r}{h}x$

• Thickness equal to differential width: $\Delta x$

The differential volume $V_{\text{disk}}$ of an individual disk is:

$V_{\text{disk}} = \pi [y(x)]^2 \Delta x = \pi \left(\frac{r}{h}x\right)^2 \Delta x = \frac{\pi r^2}{h^2} x^2 \Delta x$

Riemann Sum and Limit Formulation

Subdividing the height interval $[0, h]$ into $n$ equal subintervals of width $\Delta x = \frac{h}{n}$, the total volume is approximated by the sum of all individual disk volumes:

$V \approx \sum_{i=1}^{n} \pi \left(\frac{r}{h} x_i^*\right)^2 \Delta x$

Taking the limit as the number of slices approaches infinity ($n \to \infty$ or $\Delta x \to 0$) yields the exact volume:

$V = \lim_{n \to \infty} \sum_{i=1}^{n} \pi \left(\frac{r}{h} x_i^*\right)^2 \Delta x$

Integration and Definite Evaluation

Converting the limit of Riemann sums into a definite integral across the domain $x \in [0, h]$:

$V = \int_{0}^{h} \pi \left(\frac{r}{h}x\right)^2 dx$

Factor out constant terms from the integrand:

$V = \frac{\pi r^2}{h^2} \int_{0}^{h} x^2 dx$

Evaluate the integral using the Fundamental Theorem of Calculus:

$\int_{0}^{h} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{h} = \frac{h^3}{3} - 0 = \frac{h^3}{3}$

Multiply the result by the initial constant factors:

$V = \frac{\pi r^2}{h^2} \cdot \frac{h^3}{3} = \frac{1}{3}\pi r^2 h$

 This completes the exact proof of the cone volume formula.

Revolving Bounded Regions About the Horizontal Axis (Washer Method)

Consider the region $R$ bounded by the upper curve $y = 4\sqrt{x}$ and the lower line $y = x + 3$ between $x = 1$ and $x = 9$. Revolve region $R$ about the $x$-axis to determine the solid's total volume.

Boundary Intersections and Curve Orientation

To confirm the domain boundaries, set the two equations equal to locate intersection points:

$4\sqrt{x} = x + 3$

Square both sides of the equation:

$16x = (x + 3)^2 = x^2 + 6x + 9$

Rearrange into standard quadratic form:

$x^2 - 10x + 9 = 0$

Factor the quadratic equation:

$(x - 1)(x - 9) = 0$

Intersections occur at $x = 1$ and $x = 9$. On the interval $x \in (1, 9)$, testing $x = 4$ shows:

$y_1 = 4\sqrt{4} = 8$

$y_2 = 4 + 3 = 7$

Therefore, $4\sqrt{x} \ge x + 3$ across the entire interval $[1, 9]$.

Washer Method Setup

When revolving around the horizontal $x$-axis, vertical cross-sections form washers (annuli) perpendicular to the axis of revolution:

• Outer radius $R(x) = 4\sqrt{x}$

• Inner radius $r(x) = x + 3$

• Washer cross-sectional area: $A(x) = \pi \left( [R(x)]^2 - [r(x)]^2 \right)$

Expand the squared radii:

$[R(x)]^2 = (4\sqrt{x})^2 = 16x$

$[r(x)]^2 = (x + 3)^2 = x^2 + 6x + 9$

Subtract the inner radius squared from the outer radius squared:

$[R(x)]^2 - [r(x)]^2 = 16x - (x^2 + 6x + 9) = -x^2 + 10x - 9$

Integral Computation

Set up the definite integral for total volume:

$V = \pi \int_{1}^{9} (-x^2 + 10x - 9) dx$

Find the antiderivative:

$\int (-x^2 + 10x - 9) dx = -\frac{x^3}{3} + 5x^2 - 9x$

Evaluate at the upper limit $x = 9$:

$-\frac{9^3}{3} + 5(9)^2 - 9(9) = -\frac{729}{3} + 5(81) - 81 = -243 + 405 - 81 = 81$

Evaluate at the lower limit $x = 1$:

$-\frac{1^3}{3} + 5(1)^2 - 9(1) = -\frac{1}{3} + 5 - 9 = -\frac{1}{3} - 4 = -\frac{13}{3}$

Subtract the lower limit value from the upper limit value:

$V = \pi \left[ 81 - \left(-\frac{13}{3}\right) \right] = \pi \left( 81 + \frac{13}{3} \right) = \pi \left( \frac{243 + 13}{3} \right) = \frac{256\pi}{3}$

Solids with Known Cross-Sectional Geometries (Square Slices)

Calculate the volume of a solid whose base is bounded by lines $y = 1 - x$, $y = 2x + 5$, $x = 0$, and $x = 3$, given that every cross-section perpendicular to the $x$-axis is a square.

Determining Side Length Function

For any $x \in [0, 3]$, determine the vertical distance between the bounding functions:

$y_{\text{top}} = 2x + 5$

$y_{\text{bottom}} = 1 - x$

Since $2x + 5 > 1 - x$ for all $x \ge 0$, the side length $s(x)$ of each square cross-section is:

$s(x) = y_{\text{top}} - y_{\text{bottom}} = (2x + 5) - (1 - x) = 3x + 4$

Area Function and Integration

The cross-sectional area $A(x)$ of a square with side length $s(x)$ is:

$A(x) = [s(x)]^2 = (3x + 4)^2 = 9x^2 + 24x + 16$

Integrate $A(x)$ over the interval $x \in [0, 3]$ to find total volume:

$V = \int_{0}^{3} (9x^2 + 24x + 16) dx$

Find the antiderivative:

$\int (9x^2 + 24x + 16) dx = 3x^3 + 12x^2 + 16x$

Evaluate at $x = 3$:

$3(3)^3 + 12(3)^2 + 16(3) = 3(27) + 12(9) + 48 = 81 + 108 + 48 = 237$

Evaluate at $x = 0$:

$3(0)^3 + 12(0)^2 + 16(0) = 0$

Subtract to find the exact volume:

$V = 237$

Revolving Bounded Regions About the Vertical Axis (Cylindrical Shell Method)

Consider the region bounded by the $x$-axis ($y = 0$), lines $x = 1$ and $x = 2$, and the cubic curve $y = x^3 - x^2 + 1$. Revolve this region about the $y$-axis to form a solid resembling a cylindrical glass with a central hole in the bottom.

Analytical Failure of the Washer Method

Using the washer method for vertical axis revolutions requires horizontal slices taken perpendicular to the $y$-axis, expressing $x$ as an explicit function of $y$ ($x = f(y)$).

Attempting to solve $y = x^3 - x^2 + 1$ for $x$ requires complex cubic formulas involving radical equations. This yields impractical expressions that cannot be integrated analytically with ease. Furthermore, the inner and outer boundary boundaries change expressions across different vertical intervals of $y$. Thus, the washer method fails in practical execution.

Shell Method Structural Setup

Instead, decompose the solid using vertical cylindrical shells centered around the $y$-axis. A cylindrical shell represents the thin boundary layer of a hollow cylinder (analogous to cardboard inside paper towel rolls).

When unfolded into a flat rectangular sheet, a shell at position $x$ with thickness $\Delta x$ has:

• Circumference / Length: $2\pi \times \text{radius} = 2\pi x$

• Height: $y = x^3 - x^2 + 1$

• Thickness: $\Delta x$

The general differential volume $V_{\text{shell}}$ of a single cylindrical shell is:

$V_{\text{shell}} = 2\pi \times \text{radius} \times \text{height} \times \text{thickness} = 2\pi x y \Delta x$

Expressing purely in terms of $x$:

$V_{\text{shell}} = 2\pi x (x^3 - x^2 + 1) \Delta x$

Integration and Solution

Integrate $V_{\text{shell}}$ over the range $x \in [1, 2]$:

$V = 2\pi \int_{1}^{2} x(x^3 - x^2 + 1) dx$

Distribute $x$ through the integrand:

$V = 2\pi \int_{1}^{2} (x^4 - x^3 + x) dx$

Determine the antiderivative:

$\int (x^4 - x^3 + x) dx = \frac{x^5}{5} - \frac{x^4}{4} + \frac{x^2}{2}$

Evaluate at upper boundary $x = 2$:

$\frac{2^5}{5} - \frac{2^4}{4} + \frac{2^2}{2} = \frac{32}{5} - \frac{16}{4} + \frac{4}{2} = \frac{32}{5} - 4 + 2 = \frac{32}{5} - 2 = \frac{22}{5}$

Evaluate at lower boundary $x = 1$:

$\frac{1^5}{5} - \frac{1^4}{4} + \frac{1^2}{2} = \frac{1}{5} - \frac{1}{4} + \frac{1}{2} = \frac{4 - 5 + 10}{20} = \frac{9}{20}$

Compute the difference:

$\frac{22}{5} - \frac{9}{20} = \frac{88}{20} - \frac{9}{20} = \frac{79}{20}$

Multiply by factor $2\pi$:

$V = 2\pi \left(\frac{79}{20}\right) = \frac{79\pi}{10}$

Advanced Revolutions and Boundary-Shifted Axes

Let region $D$ be bounded by parabola $y = x^2$ and line $y = x$. The curve intersection points are derived by setting $x^2 = x \implies x(x - 1) = 0$, yielding boundaries $x = 0$ and $x = 1$. On $x \in [0, 1]$, the linear function upper-bounds the parabola ($x \ge x^2$).

Revolution About the $x$-Axis ($y = 0$)

Utilize the Washer Method with vertical cross-sections:

• Outer radius: $R(x) = x$

• Inner radius: $r(x) = x^2$

$V = \pi \int_{0}^{1} \left( x^2 - (x^2)^2 \right) dx = \pi \int_{0}^{1} (x^2 - x^4) dx$

$V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2\pi}{15}$

Revolution About the $y$-Axis ($x = 0$)

Utilize the Cylindrical Shell Method with vertical slices:

• Shell radius: $r(x) = x$

• Shell height: $h(x) = x - x^2$

$V = 2\pi \int_{0}^{1} x(x - x^2) dx = 2\pi \int_{0}^{1} (x^2 - x^3) dx$

$V = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = 2\pi \left( \frac{1}{3} - \frac{1}{4} \right) = 2\pi \left(\frac{1}{12}\right) = \frac{\pi}{6}$

Revolution About Shifted Axis $y = 3$

Utilize the Washer Method relative to horizontal axis $y = 3$

• Outer radius (distance to lower curve $y = x^2$): $R(x) = 3 - x^2$

• Inner radius (distance to upper curve $y = x$): $r(x) = 3 - x$

Calculate radius difference:

$[R(x)]^2 - [r(x)]^2 = (3 - x^2)^2 - (3 - x)^2 = (9 - 6x^2 + x^4) - (9 - 6x + x^2) = x^4 - 7x^2 + 6x$

Integrate across $x \in [0, 1]$:

$V = \pi \int_{0}^{1} (x^4 - 7x^2 + 6x) dx = \pi \left[ \frac{x^5}{5} - \frac{7x^3}{3} + 3x^2 \right]_{0}^{1}$

$V = \pi \left( \frac{1}{5} - \frac{7}{3} + 3 \right) = \pi \left( \frac{3 - 35 + 45}{15} \right) = \frac{13\pi}{15}$

Revolution About Shifted Axis $x = 5$

Utilize the Shell Method relative to vertical axis $x = 5$

• Shell radius: $r(x) = 5 - x$

• Shell height: $h(x) = x - x^2$

Product of radius and height:

$(5 - x)(x - x^2) = 5x - 5x^2 - x^2 + x^3 = x^3 - 6x^2 + 5x$

Integrate across $x \in [0, 1]$:

$V = 2\pi \int_{0}^{1} (x^3 - 6x^2 + 5x) dx = 2\pi \left[ \frac{x^4}{4} - 2x^3 + \frac{5x^2}{2} \right]_{0}^{1}$

$V = 2\pi \left( \frac{1}{4} - 2 + \frac{5}{2} \right) = 2\pi \left( \frac{1 - 8 + 10}{4} \right) = 2\pi \left(\frac{3}{4}\right) = \frac{3\pi}{2}$

Spherical Derivations via Integration (Disk vs. Shell Methods)

Derive the volume formula $V = \frac{4}{3}\pi r^3$ for a sphere with radius $r$ using two independent integration techniques.

Sphere Volume via Disk Method

Consider a circle centered at the origin: $x^2 + y^2 = r^2 \implies y^2 = r^2 - x^2$. Revolve the upper semicircle $y = \sqrt{r^2 - x^2}$ around the $x$-axis from $x = -r$ to $x = r$.

Each horizontal slice forms a disk with radius $y(x) = \sqrt{r^2 - x^2}$:

$V = \pi \int_{-r}^{r} [y(x)]^2 dx = \pi \int_{-r}^{r} (r^2 - x^2) dx$

Exploit symmetry across the $y$-axis:

$V = 2\pi \int_{0}^{r} (r^2 - x^2) dx$

Evaluate the integral:

$V = 2\pi \left[ r^2 x - \frac{x^3}{3} \right]_{0}^{r} = 2\pi \left( r^3 - \frac{r^3}{3} \right) = 2\pi \left( \frac{2r^3}{3} \right) = \frac{4}{3}\pi r^3$

Sphere Volume via Shell Method

Revolve the right vertical semicircle section defined for $x \in [0, r]$ about the $y$-axis. The full vertical height of the sphere at any distance $x$ is $2y = 2\sqrt{r^2 - x^2}$.

Using cylindrical shells:

• Shell radius: $x$

• Shell height: $2\sqrt{r^2 - x^2}$

$V = 2\pi \int_{0}^{r} x \cdot (2\sqrt{r^2 - x^2}) dx = 4\pi \int_{0}^{r} x \sqrt{r^2 - x^2} dx$

Apply substitution $u = r^2 - x^2 \implies du = -2x dx \implies x dx = -\frac{du}{2}$:

• When $x = 0 \implies u = r^2$

• When $x = r \implies u = 0$

$V = 4\pi \int_{r^2}^{0} \sqrt{u} \left(-\frac{du}{2}\right) = 2\pi \int_{0}^{r^2} u^{1/2} du$

Evaluate the integral:

$V = 2\pi \left[ \frac{2}{3}u^{3/2} \right]_{0}^{r^2} = 2\pi \left( \frac{2}{3} (r^2)^{3/2} \right) = \frac{4}{3}\pi r^3$

Cylindrical Boreholes through Spherical Solids

A cylindrical hole of radius $r$ is drilled vertically downward through the central axis of a solid sphere of radius $R$ (where $R > r$). Calculate the volume of the solid remaining after drilling.

Geometric Shell Formulation

Align the sphere's central axis along the vertical $y$-axis. The profile equation of the sphere is $x^2 + y^2 = R^2$.

Drilling a hole of radius $r$ removes all material from radial distance $x = 0$ to $x = r$. The remaining solid spans radial distances from $x = r$ to $x = R$.

Using cylindrical shells centered on the $y$-axis:

• Shell radius: $x$

• Full vertical height of sphere at radial position $x$: $h(x) = 2\sqrt{R^2 - x^2}$

• Differential shell volume element: $dV = 2\pi x (2\sqrt{R^2 - x^2}) dx = 4\pi x \sqrt{R^2 - x^2} dx$

Integral Evaluation

Set up the integral over the remaining domain $x \in [r, R]$:

$V = 4\pi \int_{r}^{R} x \sqrt{R^2 - x^2} dx$

Substitute $u = R^2 - x^2 \implies du = -2x dx \implies x dx = -\frac{du}{2}$:

• Lower limit $x = r \implies u = R^2 - r^2$

• Upper limit $x = R \implies u = R^2 - R^2 = 0$

Adjust integration boundaries:

$V = 4\pi \int_{R^2 - r^2}^{0} \sqrt{u} \left(-\frac{du}{2}\right) = 2\pi \int_{0}^{R^2 - r^2} u^{1/2} du$

Integrate:

$V = 2\pi \left[ \frac{2}{3}u^{3/2} \right]_{0}^{R^2 - r^2} = \frac{4}{3}\pi (R^2 - r^2)^{3/2}$

Volume Derivation for Square-Based Pyramids

Derive the general volume formula for a regular pyramid with a square base of side length $L$ and height $H$.

Coordinate Setup and Similar Triangles

Position the apex of the pyramid at origin $y = 0$ and extend its central height axis vertically to the base at $y = H$.

Cross-sections perpendicular to the central height axis at distance $y$ from the apex form squares of side length $s(y)$. By linear similarity of geometric profiles:

$\frac{s(y)}{L} = \frac{y}{H} \implies s(y) = \frac{L}{H}y$

Cross-Sectional Area Integration

The cross-sectional area function $A(y)$ is:

$A(y) = [s(y)]^2 = \left(\frac{L}{H}y\right)^2 = \frac{L^2}{H^2}y^2$

Integrate $A(y)$ from apex ($y = 0$) to base ($y = H$):

$V = \int_{0}^{H} \frac{L^2}{H^2} y^2 dy = \frac{L^2}{H^2} \int_{0}^{H} y^2 dy$

Evaluate the integral:

$V = \frac{L^2}{H^2} \left[ \frac{y^3}{3} \right]_{0}^{H} = \frac{L^2}{H^2} \left( \frac{H^3}{3} \right) = \frac{1}{3}L^2 H$

Additional Revolutions and Cross-Sectional Solved Problems

Cross-Sectional Solid over Semicircular Base

The base of a solid is bounded by the $x$-axis and the semicircle $y = \sqrt{16 - x^2}$. Cross-sections perpendicular to the $x$-axis are squares. Calculate the solid volume.

• Semicircle extends from $x = -4$ to $x = 4$.

• Side length of square slice: $s(x) = \sqrt{16 - x^2}$.

• Area function: $A(x) = [s(x)]^2 = 16 - x^2$

$V = \int_{-4}^{4} (16 - x^2) dx = 2 \int_{0}^{4} (16 - x^2) dx$

$V = 2 \left[ 16x - \frac{x^3}{3} \right]_{0}^{4} = 2 \left( 64 - \frac{64}{3} \right) = 2 \left( \frac{128}{3} \right) = \frac{256}{3}$

Revolution of Dual-Parabolic Region About the $y$-Axis

Consider the region bounded by $y = x^2$, $y = 1 - x^2$, and the $y$-axis for $x \ge 0$. Find the volume when revolved about the $y$-axis.

Intersection of curves: $x^2 = 1 - x^2 \implies 2x^2 = 1 \implies x = \frac{1}{\sqrt{2}}$, where $y = \frac{1}{2}$.

Washer Method Solution

Split the horizontal integration along $y$ into two distinct intervals:

1. Lower region $y \in \left[0, \frac{1}{2}\right]$: bounded by $x = \sqrt{y}$

$V_1 = \pi \int_{0}^{1/2} (\sqrt{y})^2 dy = \pi \left[ \frac{y^2}{2} \right]_{0}^{1/2} = \pi \left( \frac{1/4}{2} \right) = \frac{\pi}{8}$

2. Upper region $y \in \left[\frac{1}{2}, 1\right]$: bounded by $x = \sqrt{1 - y}$

$V_2 = \pi \int_{1/2}^{1} (\sqrt{1 - y})^2 dy = \pi \left[ y - \frac{y^2}{2} \right]_{1/2}^{1} = \pi \left[ \left(1 - \frac{1}{2}\right) - \left(\frac{1}{2} - \frac{1}{8}\right) \right] = \pi \left( \frac{1}{2} - \frac{3}{8} \right) = \frac{\pi}{8}$

Sum the partial volumes:

$V = V_1 + V_2 = \frac{\pi}{8} + \frac{\pi}{8} = \frac{\pi}{4}$

Shell Method Solution

Using vertical shells from $x = 0$ to $x = \frac{1}{\sqrt{2}}$:

• Shell radius: $x$

• Shell height: $h(x) = (1 - x^2) - x^2 = 1 - 2x^2$

$V = 2\pi \int_{0}^{1/\sqrt{2}} x(1 - 2x^2) dx = 2\pi \int_{0}^{1/\sqrt{2}} (x - 2x^3) dx$

$V = 2\pi \left[ \frac{x^2}{2} - \frac{x^4}{2} \right]_{0}^{1/\sqrt{2}} = 2\pi \left[ \frac{1/2}{2} - \frac{1/4}{2} \right] = 2\pi \left( \frac{1}{4} - \frac{1}{8} \right) = 2\pi \left(\frac{1}{8}\right) = \frac{\pi}{4}$

Summary Matrix of Formulas and Short Answers

The following table summarizes the key analytical volume expressions and answers derived across all problems: