Exercises: Volumes of revolution

The region under the curve $y = \sqrt{x}$ between $x = 0$ and $x = 4$ rotates around the $OX$ axis, as shown in the figure.

Set up the corresponding integral $V = \pi \int_{a}^{b} [f(x)]^2 \, dx$ and calculate the volume of the solid of revolution generated, expressing the result in terms of $\pi$.

The constant function $f(x) = 2$ between $x = 0$ and $x = 5$ rotates around the $OX$ axis.

1. a) Calculate the volume of the solid generated using the integral $V = \pi \int_{0}^{5} [f(x)]^2 \, dx$.
2. b) What geometric solid has been generated? Check the result using the formula for the volume of that solid.

The semicircle $y = \sqrt{9 - x^2}$ rotates around the $OX$ axis.

1. a) What geometric solid is generated and between what limits must you integrate?
2. b) Calculate its volume using the corresponding integral.
3. c) Check the result using the formula $V = \frac{4}{3} \pi r^3$.

The segment of the line $y = x$ between $x = 0$ and $x = 3$ rotates around the $OX$ axis generating a cone.

1. a) Calculate its volume using the integral $V = \pi \int_{0}^{3} x^2 \, dx$.
2. b) Identify the radius and height of the cone and check the result using the formula $V = \frac{1}{3} \pi r^2 h$.

Carlos is trying to calculate the volume of the solid of revolution generated by revolving around the $OX$-axis the region in the first quadrant bounded by the curves $f(x) = \sqrt{x}$ and $g(x) = \frac{1}{2}x$.

In his solution, he writes the following steps:

Step 1: "I set both functions equal to find the limits of integration: $\sqrt{x} = \frac{1}{2}x \implies x = 0$ and $x = 4$".

Step 2: "Since the solid has a hole inside, I subtract the functions before squaring: $V = \pi \int_0^4 \left(\sqrt{x} - \frac{1}{2}x\right)^2 dx$".

Step 3: "I expand the binomial: $\left(\sqrt{x} - \frac{1}{2}x\right)^2 = x - x^{3/2} + \frac{1}{4}x^2$".

Step 4: "I integrate and obtain the volume: $V = \pi \left[ \frac{x^2}{2} - \frac{2}{5}x^{5/2} + \frac{x^3}{12} \right]_0^4 = \frac{8\pi}{15}\ \text{u}^3$".

1. a) Identify the conceptual mistake made by Carlos in Step 2 and explain geometrically why the washer method requires subtracting the squares of the radii, $[f(x)]^2 - [g(x)]^2$, and not the square of the difference, $[f(x)-g(x)]^2$.
2. b) Correct the formulation by writing the proper integral and calculate the true volume of the solid.
3. c) Prove that if $f(x) > g(x) > 0$, it always holds that $[f(x)]^2 - [g(x)]^2 > [f(x)-g(x)]^2$, and conclude whether Carlos's calculation underestimated or overestimated the true volume.

Consider the region bounded by the curve $y = 2\sqrt{x}$, the $OX$ abscissa axis, and the vertical line $x = 4$, which revolves around the $OX$-axis generating a solid of revolution (a paraboloid).

1. a) Estimate: Before calculating any integral, justify that the solid is completely inscribed within a right cylinder of radius $R = 4$ and height $h = 4$, and that at the same time it contains inside it the right cone formed by revolving the straight line segment joining the origin $(0,0)$ to the point $(4,4)$. Calculate the exact volume of said cylinder ($V_{\text{cylinder}} = \pi R^2 h$) and of said cone ($V_{\text{cone}} = \frac{1}{3}\pi R^2 h$) to bound between which two numerical limits the volume of the solid must necessarily lie.
2. b) Calculate: Set up the definite integral $V = \pi \int_0^4 [f(x)]^2\,dx$ and calculate the exact value of the volume of the solid of revolution.
3. c) Check: Verify whether the obtained value falls within the interval estimated in section a) and determine what exact fraction of the volume of the circumscribed cylinder this paraboloid represents.

The profile of a ceramic cup is modelled by the curve $y = \sqrt{x}$, with $x$ and $y$ measured in centimetres, rotating around the $OX$ axis between $x = 0$ and $x = 9$.

1. a) Calculate the interior volume of the cup in $\text{cm}^3$, rounding to the nearest integer.
2. b) Knowing that $1\ \text{cm}^3 = 1\ \text{ml}$, what is the capacity of the cup in millilitres? Does this seem like a reasonable capacity for a cup? Comment on the result.

The region enclosed between the parabola $y = x^2$ and the line $y = x$, shaded in the figure, rotates around the $OX$ axis generating a hollow solid.

1. a) Find the intersection points of both curves.
2. b) Set up the volume as the difference of two integrals (washer method) and calculate it.

A conical laboratory funnel is obtained by revolving the line $y = 2x$, with $0 \le x \le 2$, about the $OY$-axis, as shown in the figure (measurements in decimetres).

1. a) Express $x$ in terms of $y$ and set up the integral $V = \pi \int_{c}^{d} [g(y)]^2 \, dy$ that gives the volume, stating the limits of integration.
2. b) Calculate the volume of the funnel.
3. c) Check the result using the formula for the volume of a cone, $V = \frac{1}{3} \pi r^2 h$, identifying the radius and the height of the funnel.

The nozzle of a model aircraft engine is a part whose outer wall follows the profile of the hyperbola $y = \frac{2}{x}$ between $x = 1$ and $x = 4$ (measurements in centimetres), when rotating around the $OX$ axis.

1. a) Calculate the volume of material needed to manufacture the part, assuming it is solid between the axis and the profile.
2. b) If the material has a density of $2,5\ \text{g/cm}^3$, what is the mass of the part? Round to one decimal place of a gram.

In wine aging cellars in Jerez and Rioja, an oak barrel with an exterior length of $80\ \text{cm}$ has an interior longitudinal profile modeled by the parabolic arc $y = 30 - k x^2$, with $x \in [-40, 40]$ measured in centimeters and $y$ representing the interior radius at each cross section.

The interior radius at the center of the barrel ($x = 0$) is $30\ \text{cm}$, while at the two flat ends ($x = -40$ and $x = 40$) the radius decreases to $26\ \text{cm}$.

1. a) Reasonably determine the exact value of the constant $k$ from the geometric data of the barrel.
2. b) Set up the definite integral that gives the capacity of the barrel by revolving this parabolic profile around the $OX$-axis, and calculate its exact volume in cubic centimeters (leaving the result in terms of $\pi$).
3. c) Convert the volume to liters (taking $\pi \approx 3,1416$ and rounding to tenths). Is this result consistent with the typical capacity of standard commercial barrels ($200$–$225\ \text{liters}$)? Reason whether approximating the barrel by a simple cylinder of radius equal to the mean radius ($r_m = \frac{30+26}{2} = 28\ \text{cm}$) would overestimate or underestimate the real volume obtained with the integral.

We wish to calculate the volume of the solid generated by revolving around the vertical $OY$-axis the region in the first quadrant bounded by the curve $y = \sqrt{x}$, the vertical line $x = 4$, and the horizontal $OX$-axis ($y = 0$). Three students discuss how to solve it:

• Approach A (Sofía): She observes that the maximum height on the vertical axis is $y = 2$. She proposes calculating the volume of the outer cylinder of radius $R = 4$ and height $H = 2$, and subtracting from it the volume of the interior hollow generated by revolving the curve with respect to $OY$: $V = \pi \cdot 4^2 \cdot 2 - \pi \int_0^2 [g(y)]^2 dy$, solving for $x = g(y) = y^2$.

• Approach B (Mateo): He proposes using directly the method of washers perpendicular to the $OY$-axis. Since the outer radius is fixed ($R = 4$) and the inner one varies along the curve ($r = y^2$), he sets up $V = \pi \int_0^2 \big(4^2 - (y^2)^2\big)\,dy$.

• Approach C (Lucas): He claims that there is no need to solve for $y$ and that the volume is calculated directly with $V = \pi \int_0^4 (\sqrt{x})^2 dx$.

1. a) Explain why Lucas's method (Approach C) is conceptually wrong and what geometric volume the integral he wrote actually represents.
2. b) Prove algebraically that Sofía's and Mateo's integrals lead to exactly the same expression and calculate the volume of the solid.
3. c) Critically assess which of the two correct approaches (A or B) is more advantageous when it comes to avoiding calculation errors in an exam, and justify your position.

A rugby ball is shaped like an ellipsoid of revolution: its cross-section is the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which rotates around the $OX$ axis, as shown in the figure.

1. a) Rearrange $y^2$ in the equation of the ellipse and set up the integral giving the volume of the ellipsoid.
2. b) Prove that the volume of an ellipsoid of revolution is $V = \frac{4}{3} \pi a b^2$.
3. c) A regulation ball has a length of $30\ \text{cm}$ ($a = 15\ \text{cm}$) and a maximum diameter of $18\ \text{cm}$ ($b = 9\ \text{cm}$). Calculate its volume and express it in litres, rounding to one decimal place.

The infinite region bounded by the hyperbola $y = \frac{1}{x}$, the $OX$ axis and the line $x = 1$ rotates around the $OX$ axis, generating an infinite solid known as Torricelli's trumpet, as shown in the figure.

1. a) Calculate the volume $V(b)$ of the section of the trumpet between $x = 1$ and $x = b$.
2. b) Find $\lim_{b \to +\infty} V(b)$ and interpret the result: is the volume of the full trumpet finite?
3. c) Calculate the area of the region that has rotated, that is, $\lim_{b \to +\infty} \int_{1}^{b} \frac{1}{x} dx$. Compare with the previous part and comment on the painter's paradox: you can fill the trumpet with paint, but you cannot paint its surface.

Marcos calculated the volume of the solid generated by the region under the curve $y = x^2 + 1$ between $x = -1$ and $x = 1$ when rotating around the $OX$ axis as follows:

«$V = \pi \int_{-1}^{1} (x^2 + 1) \, dx = \pi \left[ \frac{x^3}{3} + x \right]_{-1}^{1} = \pi \left( \frac{1}{3} + 1 + \frac{1}{3} + 1 \right) = \frac{8\pi}{3}\ \text{u}^3$»

1. a) Identify the error in his approach and explain what it consists of.
2. b) Calculate the volume correctly.

Consider the plane region bounded by the parabola $y = x^2$, the $OX$ axis and the line $x = 2$.

1. a) Calculate the volume of the solid generated by rotating around the $OX$ axis.
2. b) Calculate the volume of the solid generated by rotating around the $OY$ axis. (Hint: subtract from the cylinder of radius $2$ and height $4$ the solid generated by the region between the curve and the $OY$ axis.)
3. c) Although the rotating region is the same, the two volumes do not coincide. Explain why with reasons.

Consider the family of bounded regions of the plane enclosed between the parabola $y = x^2$ and the secant line $y = m x$, where the parameter $m > 0$ represents the slope of the line.

Revolving said region around the $OX$ abscissa axis generates a solid of revolution whose volume we denote by $V(m)$.

1. a) Find the intersection points between both curves in terms of $m$, set up the volume integral using the washer method, and obtain the analytical expression for $V(m)$ in terms of $\pi$ and the parameter $m$.
2. b) Prove that $V(m)$ is a function proportional to $m^5$. From this scaling property, answer without having to integrate again: if we double the slope of the line, by what factor is the volume of the new generated solid multiplied?
3. c) Solve the inverse problem: determine the exact value of the slope $m$ so that the volume of the generated solid of revolution is exactly equal to $\frac{64\pi}{15}\ \text{u}^3$.

At an AEMET weather station, a high-sensitivity rain gauge is used to record light precipitation. The collecting container has the shape of a paraboloid of revolution generated by revolving the curve $y = a x^2$ around the vertical $OY$-axis. The container has a total height of $16\ \text{cm}$ and a circular top opening of $16\ \text{cm}$ in diameter (that is, $8\ \text{cm}$ in radius).

1. a) Determine the value of parameter $a$, solve for $x = g(y)$, and calculate using integral calculus the total capacity of the container in $\text{cm}^3$ (in terms of $\pi$) and its equivalent in liters.
2. b) Derive the formula for the accumulated water volume $V(h)$ as a function of the height reached by the liquid $h$ (with $0 \le h \le 16\ \text{cm}$). From this function, calculate the exact height $h$ at which the rain gauge is at $50\,\%$ of its maximum capacity. Geometrically justify why this height is strictly greater than half the height of the container ($8\ \text{cm}$).
3. c) Calculate the instantaneous rate of change of height with respect to accumulated volume, $\frac{dh}{dV}$. Compare the value of this rate when the container is almost empty ($h \to 0$) versus when it is almost full ($h = 16\ \text{cm}$), and explain why this parabolic geometry is much more effective than that of a standard cylindrical rain gauge for detecting trace amounts of drizzle.

Unit challenge. Design a stormwater detention basin tailored to your local area by integrating GeoGebra and integral calculus.

1. Phase 1. Climate probability: Download the daily precipitation data for the past month in your region from AEMET. Model a continuous density function $f(x) = k \cdot e^{-0,2x}$ for $x \in [0, 20]$ representing litres per square metre. Analytically find $k$ to verify that $\int_0^{20} f(x)\,dx = 1$ and calculate the probability of recording more than $5\text{ mm}$ of rain.

2. Phase 2. Geometry and volume: In GeoGebra, model the basin wall using a curve $y = g(x)$ on an interval $[0, b]$. Calculate the total capacity of the basin by determining the volume of revolution generated by rotating this function around the $OX$-axis: $V = \pi \int_0^b [g(x)]^2\,dx$.

3. Phase 3. Filling and FTC: Starting from a variable inflow rate function $c(t)$, apply the Fundamental Theorem of Calculus and Barrow's rule to obtain the cumulative volume function over time via the area under the curve.

Prepare a brief technical report (maximum two pages) with the analytical workings, screenshots of the GeoGebra model, and an assessment of whether the calculated volume is adequate for the climate data.

Assessment criteria:

• Rigorous application of Barrow's rule and accurate calculation of integrals for areas and volumes of revolution.

• Mathematical justification of the properties of probability and the density function using AEMET data.

• Accurate geometric modelling in GeoGebra and interpretation of the Fundamental Theorem of Calculus.