Figure 5.5: Cylindrical Disk Slicing for Volume of Revolution about the x-axis
1. The Geometric Mechanics of Volume of Revolution
A solid of revolution is generated when a two-dimensional planar region is revolved through $360^\circ$ ($2\pi$ radians) about a fixed coplanar straight line called the axis of revolution.
In the IB DP AA HL curriculum, volumes are evaluated strictly by the Method of Slicing (Circular Disks and Annular Washers) perpendicular to the axis of rotation:
- Differential Element: Slicing a solid perpendicular to its axis produces a circular cross-section of radius $r$ and infinitesimal thickness ($dx$ or $dy$).
- Differential Volume: $dV = \text{Area of Circle} \times \text{Thickness} = \pi [r]^2 \, dx$ (or $\pi [r]^2 \, dy$).
- Riemann Sum Limit: The total volume is the limit of the sum of $n$ circular slices as thickness $\Delta x \to 0$: $$V = \lim_{n \to \infty} \sum_{i=1}^n \pi [r(x_i^*)]^2 \Delta x = \pi \int_a^b [r(x)]^2 \, dx$$
2. Revolution About Coordinate Axes
A. Rotation About the x-axis (Topic 5.11 SL & HL)
When the plane region bounded by $y = f(x)$, the $x$-axis, and vertical lines $x = a$ and $x = b$ is revolved about the $x$-axis:
$$V_x = \pi \int_a^b [f(x)]^2 \, dx = \pi \int_a^b y^2 \, dx$$Crucial Integrand Identities: Slicing produces squared functions that often require trigonometric or algebraic identities:
- $\sin^2(kx) = \frac{1 - \cos(2kx)}{2}$ and $\cos^2(kx) = \frac{1 + \cos(2kx)}{2}$
- $(e^{kx})^2 = e^{2kx}$ (remember that powers multiply: $(e^x)^2 = e^{2x}$, NOT $e^{x^2}$!)
- $\left( \frac{1}{\sqrt{x^2 + a^2}} \right)^2 = \frac{1}{x^2 + a^2} \implies \int \to \frac{1}{a}\arctan\left(\frac{x}{a}\right)$
B. Rotation About the y-axis (Topic 5.14 AHL Exclusive)
When the plane region bounded by the curve, the $y$-axis, and horizontal lines $y = c$ and $y = d$ is revolved $360^\circ$ about the $y$-axis:
$$V_y = \pi \int_c^d [g(y)]^2 \, dy = \pi \int_c^d x^2 \, dy$$Mandatory 3-Step Protocol for y-axis Rotations:
- Invert the Function: Solve the equation $y = f(x)$ explicitly for $x$ in terms of $y$: $x = g(y)$.
- Convert Limits: Ensure limits are $y$-values ($c = f(a), d = f(b)$). Never evaluate $dy$ across $x$-limits!
- Square and Integrate: Evaluate $\pi \int_c^d [g(y)]^2 dy$.
3. The Washer Method: Region Between Two Intersecting Curves
When the rotated region is bounded between an outer boundary $y_1 = f(x)$ and an inner boundary $y_2 = g(x)$ with $|f(x)| \ge |g(x)|$, cross-sections perpendicular to the axis form annular rings (washers):
$$\text{Area of Washer} = \pi R^2 - \pi r^2 = \pi \left( R^2 - r^2 \right)$$ $$V_x = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx$$Fatal Examination Trap: Difference of Squares vs Square of Difference
Never write $\pi \int (f(x) - g(x))^2 dx$! The area of a washer with outer radius $R$ and inner hole $r$ is $\pi(R^2 - r^2)$, NOT $\pi(R - r)^2$. Writing $(f - g)^2$ loses all method and accuracy marks instantly in IB markschemes!
4. Rotations About Shifted (Non-Coordinate) Lines
IB AA HL frequently examines solids revolved around horizontal lines $y = k$ or vertical lines $x = h$:
A. Rotation About a Horizontal Line $y = k$
The radius of each circular slice is the perpendicular distance from the curve to the line $y = k$:
$$r(x) = |f(x) - k| \implies V = \pi \int_a^b (f(x) - k)^2 \, dx$$If bounded between two curves $f(x)$ (outer) and $g(x)$ (inner) relative to $y = k$:
$$V = \pi \int_a^b \left( (f(x) - k)^2 - (g(x) - k)^2 \right) dx$$B. Rotation About a Vertical Line $x = h$
The radius is the horizontal distance from the curve $x = g(y)$ to the vertical axis $x = h$:
$$r(y) = |g(y) - h| \implies V = \pi \int_c^d (g(y) - h)^2 \, dy$$5. Dynamic Modeling & Fluid Rates of Change
In Paper 2 / Section B applied questions, containers (vases, goblets, bowls) are modeled by revolving curves about an axis. If fluid fills the container at rate $\frac{dV}{dt}$:
By the Fundamental Theorem of Calculus, the volume at liquid level $h$ is $V(h) = \pi \int_0^h [r(y)]^2 dy$. Differentiating with respect to $h$ yields:
$$\frac{dV}{dh} = \pi [r(h)]^2 = \text{Surface Area of Liquid}$$Connecting volume rate of change to water rising speed using the chain rule:
$$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{\pi [r(h)]^2} \frac{dV}{dt}$$Worked Examples
Worked Example 1: Trigonometric x-axis Rotation (Paper 1 Standard)
The curve $y = 2\sin(x)$ from $x = 0$ to $x = \frac{\pi}{2}$ is rotated $360^\circ$ about the $x$-axis.
Find the exact volume of the solid generated, expressing your answer in the form $k\pi^2$ where $k \in \mathbb{Q}$.
Solution:
1. State the formula for volume of revolution about the $x$-axis: [M1]
$$V = \pi \int_0^{\pi/2} y^2 \, dx = \pi \int_0^{\pi/2} [2\sin(x)]^2 \, dx = 4\pi \int_0^{\pi/2} \sin^2(x) \, dx \quad \text{[A1]}$$2. Apply the double angle identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$: [M1]
$$V = 4\pi \int_0^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx = 2\pi \int_0^{\pi/2} (1 - \cos(2x)) \, dx$$3. Integrate term-by-term: [A1]
$$V = 2\pi \left[ x - \frac{1}{2}\sin(2x) \right]_0^{\pi/2}$$4. Evaluate at the boundaries: [M1]
$$\text{Upper limit } \left(x = \frac{\pi}{2}\right): \frac{\pi}{2} - \frac{1}{2}\sin(\pi) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ $$\text{Lower limit } (x = 0): 0 - \frac{1}{2}\sin(0) = 0$$ $$V = 2\pi \left( \frac{\pi}{2} - 0 \right) = \pi^2 \quad \text{[A1]}$$Conclusion: The exact volume is $\pi^2$ (with $k = 1$).
Worked Example 2: Inverting Functions for y-axis Rotation (Paper 1 Standard)
Consider the curve $y = \ln(x - 1)$ for $x > 1$. The finite region bounded by the curve, the $y$-axis, the line $y = 0$, and the line $y = 1$ is rotated $360^\circ$ about the $y$-axis.
Determine the exact volume of the solid formed.
Solution:
1. Invert the equation to express $x$ in terms of $y$: [M1]
$$y = \ln(x - 1) \implies e^y = x - 1 \implies x = e^y + 1 \quad \text{[A1]}$$2. Set up the volume integral with respect to $y$ between $y = 0$ and $y = 1$: [M1]
$$V = \pi \int_0^1 x^2 \, dy = \pi \int_0^1 (e^y + 1)^2 \, dy$$3. Expand the binomial integrand: [M1]
$$(e^y + 1)^2 = (e^y)^2 + 2e^y + 1 = e^{2y} + 2e^y + 1$$4. Integrate term-by-term: [A1]
$$V = \pi \left[ \frac{1}{2}e^{2y} + 2e^y + y \right]_0^1$$5. Substitute limits carefully: [M1]
$$\text{Upper: } \frac{1}{2}e^2 + 2e + 1$$ $$\text{Lower: } \frac{1}{2}e^0 + 2e^0 + 0 = \frac{1}{2}(1) + 2(1) + 0 = \frac{5}{2}$$ $$V = \pi \left[ \left(\frac{1}{2}e^2 + 2e + 1\right) - \frac{5}{2} \right] = \pi \left( \frac{1}{2}e^2 + 2e - \frac{3}{2} \right) = \frac{\pi}{2}(e^2 + 4e - 3) \quad \text{[A1]}$$Worked Example 3: The Washer Method (Paper 2 Standard)
The finite region $R$ is bounded by the parabola $y = 4 - x^2$ and the line $y = x + 2$.
(a) Find the $x$-coordinates of the points of intersection of the two curves. [2 marks]
(b) The region $R$ is rotated through $2\pi$ radians about the $x$-axis. Find the exact volume of the resulting solid. [5 marks]
Solution:
(a) Points of intersection:
$$4 - x^2 = x + 2 \implies x^2 + x - 2 = 0 \quad \text{[M1]}$$ $$(x + 2)(x - 1) = 0 \implies x = -2 \quad \text{and} \quad x = 1 \quad \text{[A1]}$$(b) Washer Method Volume:
On the interval $[-2, 1]$, $4 - x^2 \ge x + 2$, so $y_{\text{outer}} = 4 - x^2$ and $y_{\text{inner}} = x + 2$. [R1]
$$V = \pi \int_{-2}^1 \left( [y_{\text{outer}}]^2 - [y_{\text{inner}}]^2 \right) dx = \pi \int_{-2}^1 \left( (4 - x^2)^2 - (x + 2)^2 \right) dx \quad \text{[M1]}$$Expand both terms: [M1]
$$(4 - x^2)^2 = 16 - 8x^2 + x^4$$ $$(x + 2)^2 = x^2 + 4x + 4$$ $$\text{Integrand } = (16 - 8x^2 + x^4) - (x^2 + 4x + 4) = x^4 - 9x^2 - 4x + 12 \quad \text{[A1]}$$Integrate: [M1]
$$V = \pi \left[ \frac{x^5}{5} - 3x^3 - 2x^2 + 12x \right]_{-2}^1$$Evaluate limits:
$$\text{At } x = 1: \frac{1}{5} - 3 - 2 + 12 = \frac{1}{5} + 7 = \frac{36}{5}$$ $$\text{At } x = -2: \frac{-32}{5} - 3(-8) - 2(4) + 12(-2) = -\frac{32}{5} + 24 - 8 - 24 = -\frac{32}{5} - 8 = -\frac{72}{5}$$ $$V = \pi \left( \frac{36}{5} - \left(-\frac{72}{5}\right) \right) = \pi \left(\frac{108}{5}\right) = \frac{108\pi}{5} \approx 67.86 \quad \text{[A1]}$$Worked Example 4: Shifted Axis Rotation & Dynamic Fluid Rate (Paper 2 Section B)
A wine glass bowl is formed by revolving the curve $y = x^2$ from $y = 0$ to $y = 4$ about the line $y = 4$, where $x \ge 0$ and dimensions are in centimetres.
(a) Find the volume of the glass bowl. [4 marks]
(b) If wine is poured into the bowl at a steady rate of $12\text{ cm}^3/\text{s}$, calculate the rate of change of the wine depth when the depth is $2\text{ cm}$. [4 marks]
Solution:
(a) Shifted Axis Volume:
The axis of rotation is the horizontal line $y = 4$. For $x \in [0, 2]$, the curve is $y = x^2$.
The radius from the line $y = 4$ down to the curve is $r(x) = 4 - x^2$. [M1]
(b) Related Rates of Change:
Let $h$ be the depth of wine measured from the bottom of the bowl ($y = 0$).
The liquid surface at height $h$ is a circular disk with radius $R(h) = x = \sqrt{h}$. [M1]
Cross-sectional surface area at depth $h$: $A(h) = \pi [R(h)]^2 = \pi (\sqrt{h})^2 = \pi h$. [A1]
By the Fundamental Theorem of Calculus, $\frac{dV}{dh} = A(h) = \pi h$.
Apply the Chain Rule: [M1]
At depth $h = 2\text{ cm}$:
$$12 = 2\pi \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{12}{2\pi} = \frac{6}{\pi}\text{ cm/s} \approx 1.91\text{ cm/s} \quad \text{[A1]}$$Practice Exercises (10 Tiered Questions)
Complete these 10 exam-style questions spanning core skills, standard exam problems, and advanced HL challenges. Open each question to inspect the full step-by-step IB markscheme.
Find the volume of the solid generated when the curve $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is revolved $360^\circ$ about the $x$-axis.
Show Full Solution
Formula for rotation about the $x$-axis: $V = \pi \int_a^b y^2 \, dx$. [M1]
$$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx \quad \text{[A1]}$$ $$V = \pi \left[ \frac{1}{2}x^2 \right]_0^4 = \pi \left( \frac{16}{2} - 0 \right) = 8\pi \quad \text{[A1]}$$Find the volume of the solid generated by revolving the region bounded by $y = e^x$, the lines $x = 0$, $x = 2$, and the $x$-axis about the $x$-axis.
Show Full Solution
Set up the integral: $V = \pi \int_0^2 (e^x)^2 \, dx$. [M1]
Using index laws $(e^x)^2 = e^{2x}$: [A1]
$$V = \pi \int_0^2 e^{2x} \, dx = \pi \left[ \frac{1}{2}e^{2x} \right]_0^2 = \frac{\pi}{2}(e^4 - e^0) = \frac{\pi}{2}(e^4 - 1) \quad \text{[A1]}$$Examiner Pitfall: Do not write $e^{x^2}$. Powers multiply: $(e^x)^2 = e^{2x}$.
The region enclosed by $y = x^2$, the $y$-axis, and the line $y = 4$ is revolved $360^\circ$ about the $y$-axis. Calculate the volume.
Show Full Solution
Express $x^2$ in terms of $y$: $y = x^2 \implies x^2 = y$. [M1]
Set up the integral with respect to $y$ from $y = 0$ to $y = 4$: [M1]
$$V = \pi \int_0^4 x^2 \, dy = \pi \int_0^4 y \, dy$$ $$V = \pi \left[ \frac{1}{2}y^2 \right]_0^4 = \pi \left( \frac{16}{2} \right) = 8\pi \quad \text{[A1]}$$Find the volume formed when the region bounded by $y = \sin x$ from $x = 0$ to $x = \pi$ and the $x$-axis is rotated $360^\circ$ about the $x$-axis.
Show Full Solution
Set up the volume integral: $V = \pi \int_0^\pi \sin^2 x \, dx$. [M1]
Substitute double angle identity $\sin^2 x = \frac{1 - \cos 2x}{2}$: [M1]
$$V = \pi \int_0^\pi \frac{1 - \cos 2x}{2} \, dx = \frac{\pi}{2} \left[ x - \frac{1}{2}\sin 2x \right]_0^\pi \quad \text{[A1]}$$Substitute limits:
$$V = \frac{\pi}{2} \left( \pi - \frac{1}{2}\sin(2\pi) - (0 - 0) \right) = \frac{\pi}{2}(\pi - 0) = \frac{\pi^2}{2} \quad \text{[A1]}$$The region between $y = x$ and $y = x^2$ is revolved $360^\circ$ about the $x$-axis. Find the volume of the resulting solid.
Show Full Solution
Points of intersection: $x = x^2 \implies x(x - 1) = 0 \implies x = 0, 1$. [A1]
On $[0, 1]$, $x \ge x^2$, so $y_{\text{outer}} = x$ and $y_{\text{inner}} = x^2$. [R1]
$$V = \pi \int_0^1 \left( [x]^2 - [x^2]^2 \right) dx = \pi \int_0^1 (x^2 - x^4) \, dx \quad \text{[M1]}$$ $$V = \pi \left[ \frac{1}{3}x^3 - \frac{1}{5}x^5 \right]_0^1 = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2\pi}{15} \quad \text{[A1]}$$Find the volume generated when the curve $y = \ln x$, from $y = 0$ to $y = 2$, is rotated $360^\circ$ about the $y$-axis.
Show Full Solution
Invert to express $x$ in terms of $y$: $y = \ln x \implies x = e^y$. [M1]
Set up integral with respect to $y$: [M1]
$$V = \pi \int_0^2 x^2 \, dy = \pi \int_0^2 (e^y)^2 \, dy = \pi \int_0^2 e^{2y} \, dy$$ $$V = \pi \left[ \frac{1}{2}e^{2y} \right]_0^2 = \frac{\pi}{2}(e^4 - e^0) = \frac{\pi}{2}(e^4 - 1) \quad \text{[A1]}$$The region enclosed by $y = \frac{1}{\sqrt{x^2 + 1}}$, the $x$-axis, $x = 0$, and $x = 1$ is rotated $360^\circ$ about the $x$-axis. Find the volume.
Show Full Solution
Square the function: $y^2 = \left( \frac{1}{\sqrt{x^2 + 1}} \right)^2 = \frac{1}{x^2 + 1}$. [M1]
$$V = \pi \int_0^1 y^2 \, dx = \pi \int_0^1 \frac{1}{x^2 + 1} \, dx \quad \text{[A1]}$$Recognize standard arctan antiderivative: [M1]
$$V = \pi [\arctan x]_0^1 = \pi (\arctan(1) - \arctan(0)) = \pi \left( \frac{\pi}{4} - 0 \right) = \frac{\pi^2}{4} \quad \text{[A1]}$$Find the volume generated by rotating the region enclosed by $y = x^2$ and $y = 4$ about the line $y = 4$.
Show Full Solution
The line of rotation is $y = 4$. Radius from axis to curve: $r(x) = 4 - x^2$. [M1]
Limits: $x^2 = 4 \implies x = -2$ to $x = 2$. By symmetry: $2\pi \int_0^2$. [R1]
$$V = \pi \int_{-2}^2 (4 - x^2)^2 \, dx = 2\pi \int_0^2 (16 - 8x^2 + x^4) \, dx \quad \text{[M1]}$$ $$V = 2\pi \left[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \right]_0^2 = 2\pi \left( 32 - \frac{64}{3} + \frac{32}{5} \right) \quad \text{[A1]}$$ $$32 - \frac{64}{3} + \frac{32}{5} = \frac{480 - 320 + 96}{15} = \frac{256}{15} \implies V = \frac{512\pi}{15} \quad \text{[A1]}$$A glass goblet is modeled by revolving $y = 2 + \sqrt{x}$ from $x = 0$ to $x = 9$ about the $x$-axis (all dimensions in cm).
(a) Find the total volume of the solid in $\text{cm}^3$.
(b) Water fills the goblet at $10\text{ cm}^3/\text{s}$. Find the rate of change of depth $h$ when $h = 4\text{ cm}$.
Show Full Solution
(a) Volume of solid:
$$V = \pi \int_0^9 (2 + \sqrt{x})^2 \, dx = \pi \int_0^9 (4 + 4x^{1/2} + x) \, dx \quad \text{[M1]}$$ $$V = \pi \left[ 4x + \frac{8}{3}x^{3/2} + \frac{x^2}{2} \right]_0^9 = \pi \left( 36 + \frac{8}{3}(27) + \frac{81}{2} \right) \quad \text{[A1]}$$ $$V = \pi (36 + 72 + 40.5) = 148.5\pi = \frac{297\pi}{2}\text{ cm}^3 \quad \text{[A1]}$$(b) Related rates for depth $h$:
By the Fundamental Theorem of Calculus: $\frac{dV}{dh} = \pi [2 + \sqrt{h}]^2$. [M1]
At $h = 4$: $\frac{dV}{dh} = \pi (2 + \sqrt{4})^2 = \pi (4)^2 = 16\pi$. [A1]
Using chain rule $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$: [M1]
$$10 = 16\pi \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{10}{16\pi} = \frac{5}{8\pi}\text{ cm/s} \approx 0.199\text{ cm/s} \quad \text{[A1]}$$A circle of radius $r$ centered at $(R, 0)$ where $R > r$ has equation $(x - R)^2 + y^2 = r^2$. Rotating this circle $360^\circ$ about the $y$-axis generates a torus. Prove using calculus that the volume of the torus is $V = 2\pi^2 R r^2$.
Show Full Solution
Solve circle equation for $x$: $(x - R)^2 = r^2 - y^2 \implies x = R \pm \sqrt{r^2 - y^2}$. [M1]
Outer radius: $x_2 = R + \sqrt{r^2 - y^2}$; Inner radius: $x_1 = R - \sqrt{r^2 - y^2}$.
Washer difference of squares: [M1]
$$x_2^2 - x_1^2 = (x_2 - x_1)(x_2 + x_1) = \left(2\sqrt{r^2 - y^2}\right)(2R) = 4R\sqrt{r^2 - y^2} \quad \text{[A1]}$$Volume integral across $y \in [-r, r]$: [M1]
$$V = \pi \int_{-r}^r (x_2^2 - x_1^2) \, dy = 4\pi R \int_{-r}^r \sqrt{r^2 - y^2} \, dy$$Geometric evaluation: $\int_{-r}^r \sqrt{r^2 - y^2} dy$ is the area of a semicircle of radius $r$, which is $\frac{1}{2}\pi r^2$: [R1]
$$V = 4\pi R \left( \frac{1}{2}\pi r^2 \right) = 2\pi^2 R r^2 \quad \text{[AG]}$$Formula Booklet References
Syllabus 5.11: Revolution about the x-axis
$$V = \pi \int_a^b y^2 \, dx = \pi \int_a^b [f(x)]^2 \, dx$$Syllabus 5.14 AHL: Revolution about the y-axis
$$V = \pi \int_c^d x^2 \, dy = \pi \int_c^d [g(y)]^2 \, dy$$Condition: The equation must be rearranged to express $x$ as a function of $y$ ($x = g(y)$), and limits $c$ and $d$ must be $y$-coordinates.
Key Formulas & Identities
1. The Washer Method (Hollow Solids)
• About $x$-axis: $V = \pi \int_a^b \left( [y_{\text{outer}}]^2 - [y_{\text{inner}}]^2 \right) dx$
• About $y$-axis: $V = \pi \int_c^d \left( [x_{\text{outer}}]^2 - [x_{\text{inner}}]^2 \right) dy$
2. Shifted Axes of Revolution
• About horizontal line $y = k$: $V = \pi \int_a^b (f(x) - k)^2 \, dx$
• About vertical line $x = h$: $V = \pi \int_c^d (g(y) - h)^2 \, dy$
3. Dynamic Fluid Rates
$$\frac{dV}{dh} = \pi [r(h)]^2 \quad \text{and} \quad \frac{dh}{dt} = \frac{1}{\pi [r(h)]^2} \frac{dV}{dt}$$Common Exam Pitfalls
Forgetting the Multiplier $\pi$
The most pervasive error across all IB papers is omitting the constant $\pi$. Double-check that your volume expression begins with $\pi$.
The Washer Square Error: $(f - g)^2$ vs $(f^2 - g^2)$
Never square the difference of functions! The area of a washer ring is $\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$. Calculating $[f(x) - g(x)]^2$ results in an immediate loss of all method marks.
Mixing $x$ and $y$ in y-axis Rotations
When rotating about the $y$-axis, the entire integral must be in terms of $y$. Integrating $x^2$ with respect to $y$ requires replacing $x^2$ with $[g(y)]^2$, and the integration limits must be $y$-coordinates.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
What is the volume obtained by rotating the line $y = 2x$ from $x = 0$ to $x = 3$ about the $x$-axis?
Which integral calculates the volume generated by revolving the region between $y = \sqrt{x}$ and $y = x$ about the $x$-axis?
Revolving $y = x^3$ (for $0 \le x \le 2$) about the $y$-axis gives which integral for volume?
When the region under $y = f(x)$ from $x=a$ to $x=b$ is revolved about the horizontal line $y = -2$, the radius of each disk is:
Section B: Examination-Style Free Response Questions
Let $R$ be the region enclosed by $y = e^x$, the line $x = 1$, and both coordinate axes.
(a) Find the volume $V_x$ when $R$ is revolved $360^\circ$ about the $x$-axis. [3 marks]
(b) Find the volume $V_y$ when $R$ is revolved $360^\circ$ about the $y$-axis. [4 marks]
View Official IB Markscheme
(a) Volume around $x$-axis:
$$V_x = \pi \int_0^1 (e^x)^2 dx = \pi \int_0^1 e^{2x} dx \quad \text{[M1]}$$ $$V_x = \pi \left[ \frac{1}{2}e^{2x} \right]_0^1 = \frac{\pi}{2}(e^2 - 1) \quad \text{[A1][A1]}$$(b) Volume around $y$-axis:
The region spans $y$ from $0$ to $e$. From $y=0$ to $y=1$, it is a cylinder of radius $x=1$. From $y=1$ to $y=e$, the curve bounds it with outer radius $1$ and inner radius $x = \ln y$ [M1]:
$$V_y = \pi \int_0^1 (1)^2 dy + \pi \int_1^e \left( 1^2 - (\ln y)^2 \right) dy \quad \text{[M1]}$$ $$V_1 = \pi(1) = \pi$$Using $\int (\ln y)^2 dy = y(\ln y)^2 - 2y\ln y + 2y$ [M1]:
$$\int_1^e (\ln y)^2 dy = [e(1) - 2e(1) + 2e] - [0 - 0 + 2] = e - 2$$ $$\pi \int_1^e (1 - (\ln y)^2) dy = \pi [ (e - 1) - (e - 2) ] = \pi(1) = \pi$$ $$V_y = \pi + \pi = 2\pi \quad \text{[A1]}$$An architectural dome has the shape of a solid formed by rotating $y = 6 - \frac{x^2}{6}$ about the $y$-axis, where $x, y$ are measured in metres and $y \ge 0$.
(a) Find the maximum height and the ground radius of the dome. [2 marks]
(b) Calculate the enclosed volume of the dome in $\text{m}^3$. [4 marks]
View Official IB Markscheme
(a) Dimensions:
At $x=0$, maximum height is $y = 6\text{ m}$ [A1].
At ground level $y=0$: $6 - \frac{x^2}{6} = 0 \implies x^2 = 36 \implies x = 6\text{ m}$ ground radius [A1].
(b) Volume calculation:
Express $x^2$ in terms of $y$:
$$\frac{x^2}{6} = 6 - y \implies x^2 = 6(6 - y) = 36 - 6y \quad \text{[M1]}$$Integrate with respect to $y$ from $y = 0$ to $y = 6$ [M1]:
$$V = \pi \int_0^6 x^2 dy = \pi \int_0^6 (36 - 6y) dy \quad \text{[A1]}$$ $$V = \pi \left[ 36y - 3y^2 \right]_0^6 = \pi [ 36(6) - 3(36) ] = \pi [ 216 - 108 ] = 108\pi\text{ m}^3 \quad \text{[A1]}$$ $$\approx 339.3\text{ m}^3$$