Solved Exercises. Determining Volumes

The man who asks a question is a fool for a minute, the man who does not ask is a fool for life, Confucius

Calculating volumes

Calculating volumes with integrals is tricky and involves a variety of techniques such as slicing, the shell method, and the washer method.

• General Slicing Method: Used when the base is a flat region and cross-sections (like squares, triangles, or semicircles) are built on top of it. The volume is the integral of the cross-sectional area: $\boxed{V = \int_{a}^{b} A(x) \,dx}$
• Disk / Washer Method: Used for solids of revolution. It slices the solid perpendicular to the axis of rotation, creating circular disks (or washers, if there’s a hole). $V = \pi \int_{a}^{b} \big(R(x)^2 - r(x)^2\big) \,dx$
• Shell Method: Also for solids of revolution, but it slices parallel to the axis of rotation, creating nested cylindrical shells. This is often easier when solving for the radius in terms of the other variable is difficult. $V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \,dx$

Solved exercises

The Cylindrical Shell Method

The shell method is highly effective when rotating a region around the y-axis but integrating with respect to $x$. The volume formula is $\boxed{V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \,dx}$.

The Witch’s Cauldron (I know, scary, ha ha ha!)

Calculate the volume generated by the curve $y = x^2$ when it is revolved around the y-axis and bounded above by $y = a$ (Figure 1.d).

To calculate the volume generated by revolving the curve $y = x^2$ around the y-axis, the method of cylindrical shells is a great choice.

1. Setup the Shell:
   We are slicing the volume into infinitely thin vertical cylinders (shells).
   Thickness: $dx$
   Radius ($r$): The distance from the y-axis to the shell is $x$.
   Height ($h$): The top of the region is bounded by $y = a$ and the bottom by the curve $y = x^2$. Therefore, the height is $a - x^2$.

2. Find the Limits of Integration:
   The region starts at the y-axis ($x=0$) and extends to the intersection of $y = x^2$ and $y = a$, which occurs at $x = \sqrt{a}$.

3. Integrate:
   $V = \int_{0}^{\sqrt{a}} 2\pi x (a - x^2) \,dx = 2\pi \int_{0}^{\sqrt{a}} (ax - x^3) \,dx = 2\pi \left[ \frac{ax^2}{2} - \frac{x^4}{4} \right]_{0}^{\sqrt{a}}$
   Evaluate the antiderivative at the upper and lower bounds: $V = 2\pi \left( \frac{a(\sqrt{a})^2}{2} - \frac{(\sqrt{a})^4}{4} \right) = 2\pi \left( \frac{a^2}{2} - \frac{a^2}{4} \right)$
   $V = 2\pi \left( \frac{a^2}{4} \right) = \frac{\pi a^2}{2}$

Polynomial Region Revolved about the y-axis

Find the volume of the solid generated when the region bounded by $y = x^2+1$, $y = -x+1$, and $x = 1$ is revolved about the y-axis (Figure $\alpha$) (Figure $\alpha$).

Because it is revolved about the y-axis, we use cylindrical shells.

1. Find Points of Intersection:
   Find where the upper and lower curves meet to establish bounds.
   $x^2 + 1 = -x + 1 \implies x^2 + x = 0 \implies x(x + 1) = 0$
   The curves intersect at $x = 0$ and $x = -1$. Since the region is bounded on the right by $x = 1$, we evaluate from $x = 0$ to $x = 1$.

2. Setup the Shell:
   Radius ($r$): $x$
   Height ($h$): The upper curve minus the lower curve: $(x^2 + 1) - (-x + 1) = x^2 + x$

3. Integrate:
   $V = 2\pi \int_{0}^{1} x(x^2 + x) \,dx = 2\pi \int_{0}^{1} (x^3 + x^2) \,dx = 2\pi \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_{0}^{1}$
   $V = 2\pi \left( \frac{1}{4} + \frac{1}{3} \right) = 2\pi \left( \frac{7}{12} \right) = \frac{7\pi}{6}$

The Disk and Washer Methods

When revolving around the x-axis and integrating with respect to $x$, we use disks ($V = \pi \int R^2 \,dx$) or washers ($V = \pi \int (R^2 - r^2) \,dx$). If revolving around the y-axis and integrating with respect to $y$, the same formulas apply, but functions must be written as $x = f(y)$.

Simple Disk Method

Find the volume of the solid of revolution bounded by the graph of $f(x) = \sqrt{x}$, $x = 1$ and $x = 4$, and rotated about the x-axis (Figure B).

The solid of revolution is formed by revolving the region around the x-axis. The cross-sections are solid circles.

• Radius: $r = f(x) = \sqrt{x}$
• Area: $A(x) = \pi r^2 = \pi(\sqrt{x})^2 = \pi x$
• Integrate: $V = \int_{1}^{4} \pi x \,dx = \pi \left[ \frac{x^2}{2} \right]_{1}^{4}$

$V = \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \frac{15\pi}{2}$

Washer Method with Logarithmic Bounds

Find the volume of the solid generated when the region bounded by $y = e^{\frac{x}{2}}$, $y = e^{-\frac{x}{2}}$, $x = \ln(2)$, and $x = \ln(3)$ is revolved about the x-axis (Figure 5).

We have a hollow center, so we use the Washer Method.

• Outer Radius ($R$): $y_1 = e^{\frac{x}{2}}$
• Inner Radius ($r$): $y_2 = e^{-\frac{x}{2}}$

Integrate: $V = \pi \int_{\ln(2)}^{\ln(3)} \left( \left( e^{\frac{x}{2}} \right)^2 - \left( e^{-\frac{x}{2}} \right)^2 \right) \,dx = \pi \int_{\ln(2)}^{\ln(3)} (e^x - e^{-x}) \,dx$

$V = \pi \left[ e^x + e^{-x} \right]_{\ln(2)}^{\ln(3)} = \pi \left( (e^{\ln(3)} + e^{-\ln(3)}) - (e^{\ln(2)} + e^{-\ln(2)}) \right)$

Recall that $e^{-\ln(a)} = e^{\ln(a^{-1})} = \frac{1}{a}$. Therefore, $V = \pi \left( \left( 3 + \frac{1}{3} \right) - \left( 2 + \frac{1}{2} \right) \right) = \pi \left( 1 + \frac{1}{3} - \frac{1}{2} \right) = \pi \left( \frac{6}{6} + \frac{2}{6} - \frac{3}{6} \right) = \frac{5\pi}{6}$

Washer Method: Two Intersecting Functions

Find the volume of revolution if the region enclosed in the first quadrant by $y = x$ and $y = \sqrt{x}$ is rotated about the x-axis (Figure $\beta$).

Find Intersections: $x = \sqrt{x} \implies x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0$

The limits are $x = 0$ and $x = 1$. In the first quadrant between $0$ and $1$, $\sqrt{x} \ge x$, making it the outer radius.

Integrate: $V = \pi \int_{0}^{1} \left( (\sqrt{x})^2 - (x)^2 \right) \,dx = \pi \int_{0}^{1} (x - x^2) \,dx$

$V = \pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \pi \left( \frac{1}{2} - \frac{1}{3} \right) = \pi \left( \frac{3 - 2}{6} \right) = \frac{\pi}{6} \approx 0.5236 \text{ units}^3$

Washer Method Integrating with respect to y

Find the volume of the solid generated when the region bounded by $y = x$, $y = 2x$, and $y = 6$ is revolved about the y-axis (Figure $\epsilon$).

Since we are revolving around the y-axis, we can slice horizontally. This requires writing our functions in terms of $y$:

• Inner curve: $y = 2x \implies x = \frac{y}{2}$
• Outer curve: $y = x \implies x = y$
• Integrate from $y=0$ to $y=6$: $V = \pi \int_{0}^{6} \left( (y)^2 - \left(\frac{y}{2}\right)^2 \right) \,dy = \pi \int_{0}^{6} \left( y^2 - \frac{y^2}{4} \right) \,dy = \pi \int_{0}^{6} \frac{3}{4}y^2 \,dy$

$V = \frac{3\pi}{4} \left[ \frac{y^3}{3} \right]_{0}^{6} = \frac{3\pi}{4} \left( \frac{6^3}{3} \right) = \frac{3\pi}{4} (72) = 54\pi \approx 169.646 \text{ units}^3$

General Slicing (Known Cross-Sections)

When dealing with a base on the xy-plane and geometric cross-sections popping out of the plane, the volume is the integral of the cross-sectional area: $V = \int A(x) \,dx$.

Square Cross-Sections

Find the volume of the solid whose base is bounded by the curves $y = x^2$ and $y = 2 - x^2$, and whose cross sections perpendicular to the x-axis are squares (Figure C).

1. Find Intersections: $x^2 = 2 - x^2 \implies 2x^2 - 2 = 0 \implies 2(x^2 - 1) = 0 \implies x = \pm 1$

2. Find Area of Cross-Section ($A(x)$): The length of the side of each square is the upper curve minus the lower curve: $s = (2 - x^2) - x^2 = 2 - 2x^2$.
   Therefore, the area of each square is $A(x) = s^2 = (2 - 2x^2)^2 = 4 - 8x^2 + 4x^4$

3. Integrate (Utilizing Even Symmetry): $V = \int_{-1}^{1} (4 - 8x^2 + 4x^4) \,dx = 2 \int_{0}^{1} (4 - 8x^2 + 4x^4) \,dx = 2 \left[ 4x - \frac{8}{3}x^3 + \frac{4}{5}x^5 \right]_{0}^{1}$
   $V = 2 \left( 4 - \frac{8}{3} + \frac{4}{5} \right) = 2 \left( \frac{60}{15} - \frac{40}{15} + \frac{12}{15} \right) = 2 \left( \frac{32}{15} \right) = \frac{64}{15}$

• Calculate the volume of the solid whose base is the region bounded by the curve $y = \sqrt{\cos(x)}$ and the x-axis on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and whose cross sections perpendicular to the x-axis are isosceles right triangles with a horizontal leg in the xy-plane and a vertical leg parallel to the z-axis (Figure D).

1. Recall. An isosceles right triangle is a special three-sided shape with one 90-degree angle and two equal 45-degree angles. Its two shorter sides are equal in length, and its longest side (the hypotenuse) is equal to the length of a short side multiplied by the square root of 2.

2. Find Area of Cross-Section ($A(x)$): An isosceles right triangle with leg length $L$ has an area of $\frac{1}{2}L^2$. The leg in the xy-plane stretches from the x-axis to the curve, so $L = \sqrt{\cos(x)}$.
   Therefore, $A(x) = \frac{1}{2} \left( \sqrt{\cos(x)} \right)^2 = \frac{1}{2} \cos(x)$

3. Integrate (Utilizing Even Symmetry): $V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} A(x) \,dx = 2 \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cos(x) \,dx$
   $V = \int_{0}^{\frac{\pi}{2}} \cos(x) \,dx = \left[ \sin(x) \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.